The RS Aggarwal Class 10 solutions of maths for Chapter 10 Ex 10.8 are prepared with an aim to help students facilitate their preparation for the exam. We at BYJU’S provide accurate solutions of the RS Aggarwal Class 10 Maths textbook and explains each and every topic in an easy manner and give detailed solutions to difficult questions. So, by referring to these solutions, you can improve your performance and score better marks.

The solutions are prepared by highly skilled subject experts are in accordance with the latest syllabus of the CBSE. Students of Class 10 are advised to practice from RS Aggarwal Solutions Class 10 Chapter 10 – Quadratic Equations Ex 10.8Â so that they can solve all the questions that will be asked in the board exams.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 10â€“ Quadratic Equations 10H (10.8)

**Question 1:**

**Which of the following is a quadratic equation?**

**(a) \(x^{2}-3\sqrt{x}+2=0\)**

**(b) \(x+\frac{1}{x}=x^{2}\)**

**(c) \(x^{2}+\frac{1}{x^{2}}=5\)**

**(d) \(2x^{2}-5x=(x-1)^{2}\)**

**Solution:**

(d) \(2x^{2}-5x=(x-1)^{2}\)

A quadratic equation is the equation with degree 2.

Because, 2x^{2} â€“ 5x = (x â€“ 1)^{2}

\(\Rightarrow\) 2x^{2} â€“ 5x = x^{2} â€“ 2x + 1

\(\Rightarrow\) 2x^{2} â€“ 5x â€“ x^{2} + 2x â€“ 1 = 0

\(\Rightarrow\) x^{2} â€“ 3x â€“ 1 = 0, which is a quadratic equation

**Question 2:**

**Which of the following is a quadratic equation?**

**(a) (x ^{2} + 1) = (2 â€“ x)^{2} + 3**

**(b) x ^{3} â€“ x^{2} = (x â€“ 1)^{3}**

**(c) 2x ^{2} + 3 = (5 + x)(2x â€“ 3)**

**(d) none of these**

**Solution:**

(b) x^{3} â€“ x^{2} = (x â€“ 1)^{3}

Because, x^{3} â€“ x^{2} = (x â€“ 1)^{3}

\(\Rightarrow\) x^{3} â€“ x^{2} = x^{3} â€“ 3x^{2} + 3x â€“ 1

\(\Rightarrow\) 2x^{2} â€“ 3x + 1 = 0, which is a quadratic equation

**Question 3:**

**Which of the following is not a quadratic equation?**

**(a) 3x â€“ x ^{2} = x^{2} + 5**

**(b) (x + 2) ^{2} = 2(x^{2} â€“ 5)**

**(c) \((\sqrt{2}x+3)^{2}=2x^{2}+6\)**

**(d) (x â€“ 1) ^{2} = 3x^{2} + x â€“ 2**

**Solution:**

(c) \((\sqrt{2}x+3)^{2}=2x^{2}+6\)

Because, \((\sqrt{2}x+3)^{2}=2x^{2}+6\)

\(\Rightarrow\) \(2x^{2}+9+6\sqrt{2}x=2x^{2}+6\)

\(\Rightarrow\) \(6\sqrt{2}x+3=0\), which is not a quadratic equation

**Question 4:**

**If x = 3 is a solution of the equation 3x ^{2} + (k â€“ 1)x + 9 = 0, then k = ?**

**(a) 11**

**(b) -11**

**(c) 13**

**(d) -13**

**Solution:**

(b) -11

It is given that x = 3 is a solution of 3x^{2} + (k â€“ 1)x + 9 = 0; therefore, we have:

3(3)^{2} + (k â€“ 1) x 3 + 9 = 0

\(\Rightarrow\) 27 + 3(k â€“ 1) + 9 = 0

\(\Rightarrow\) 3(k â€“ 1) = -36

\(\Rightarrow\) (k â€“ 1) = -12

\(\Rightarrow\) k = -11

**Question 5:**

**If one root of the equation 2x ^{2} + ax + 6 = 0 is 2 then a = ?**

**(a) 7**

**(b) -7**

**(c) \(\frac{7}{2}\)**

**(d) \(\frac{-7}{2}\)**

**Solution:**

(b) -7

It is given that one root of the equation 2x^{2} + ax + 6 = 0 is 2.

Therefore, 2 x 2^{2} + a x 2 + 6 = 0

\(\Rightarrow\) 2a + 14 = 0

\(\Rightarrow\) a = -7

**Question 6:**

**The sum of the roots of the equation x ^{2} â€“ 6x + 2 = 0 is**

**(a) 2**

**(b) -2**

**(c) 6**

**(d) -6**

**Solution:**

(c) 6

Sum of the roots of the equation x^{2} â€“ 6x + 2 = 0 is

\(\alpha +\beta =\frac{-b}{a}=\frac{-(-6)}{1}=6\), where \(\alpha\) and \(\beta\) are the roots of the equation.

**Question 7:**

**If the product of the roots of the equation x ^{2} â€“ 3x + k = 10 is -2 then the value of k is**

**(a) -2**

**(b) -8**

**(c) 8**

**(d) 12**

**Solution:**

(c) 8

It is given that the product of the roots of the equation x^{2} â€“ 3x + k = 10 is -2.

The equation can be rewritten as:

x^{2} â€“ 3x + (k â€“ 10) = 0

Product of the roots of a quadratic equation = \(\frac{c}{a}\)

\(\Rightarrow\) \(\frac{c}{a}\) = -2

\(\Rightarrow\) \(\frac{(k-10)}{1}\) = -2

\(\Rightarrow\) k = 8

**Question 8:**

**The ratio of the sum and product of the roots of the equation 7x ^{2} â€“ 12x + 18 = 0 is**

**(a) 7 : 12**

**(b) 7 : 18**

**(c) 3 : 2**

**(d) 2 : 3**

**Solution:**

(c) 2 : 3

Given:

7x^{2} â€“ 12 x + 18 = 0

Therefore, \(\alpha +\beta =\frac{12}{7}\) and \(\alpha \beta =\frac{18}{7}\), where \(\alpha\) and \(\beta\) are the roots of the equation

Therefore, Ratio of the sum and product of the roots = \(\frac{12}{7}:\frac{18}{7}\)

= 12 : 18

= 2 : 3

**Question 9:**

**If one root of the equation 3x ^{2} â€“ 10x + 3 = 0 is \(\frac{1}{3}\) then the other root is**

**(a) \(\frac{-1}{3}\)**

**(b) \(\frac{1}{3}\)**

**(c) -3**

**(d) 3**

**Solution:**

(d) 3

Given:

3x^{2} â€“ 10x + 3 = 0

One root of the equation is \(\frac{1}{3}\).

Let the other root be \(\alpha\).

We know that:

Product of the roots = \(\frac{c}{a}\)

\(\Rightarrow\) \(\frac{1}{3}\times \alpha =\frac{3}{3}\)

\(\Rightarrow\) \(\alpha\) = 3

**Question 10:**

**If one root of 5x ^{2} + 13x + k = 0 be the reciprocal of the other root then the value of k is**

**(a) 0**

**(b) 1**

**(c) 2**

**(d) 5**

**Solution:**

(d) 5

Let the roots of the equation (5x^{2} + 13x + k = 0) be \(\alpha\) and \(\frac{1}{\alpha }\).

Therefore, Product of the roots = \(\frac{c}{a}\)

\(\Rightarrow\) \(\alpha \times \frac{1}{\alpha } =\frac{k}{5}\)

\(\Rightarrow\) 1 = \(\frac{k}{5}\)

\(\Rightarrow\) k = 5

**Question 11:**

**If the sum of the roots of the equation kx ^{2} + 2x + 3k = 0 is equal to their product then the value of k is**

**(a) \(\frac{1}{3}\)**

**(b) \(\frac{-1}{3}\)**

**(c) \(\frac{2}{3}\)**

**(d) \(\frac{-2}{3}\)**

**Solution:**

(d) \(\frac{-2}{3}\)

Given:

kx^{2} + 2x + 3k = 0

Sum of the roots = Product of the roots

\(\Rightarrow\) \(\frac{-2}{k}=\frac{3k}{k}\)

\(\Rightarrow\) 3k = -2

\(\Rightarrow\) k = \(\frac{-2}{3}\)

**Question 12:**

**The roots of a quadratic equation are 5 and -2. Then, the equation is**

**(a) x ^{2} â€“ 3x + 10 = 0**

**(b) x ^{2} â€“ 3x â€“ 10 = 0**

**(c) x ^{2} + 3x â€“ 10 = 0**

**(d) x ^{2} + 3x + 10 = 0**

**Solution:**

(b) x^{2} â€“ 3x â€“ 10 = 0

It is given that the roots of the quadratic equation are 5 and -2.

Then, the equation is:

x^{2} â€“ (5 â€“ 2)x + 5 x (-2) = 0

\(\Rightarrow\) x^{2} â€“ 3x â€“ 10 = 0

**Question 13:**

**If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is**

**(a) x ^{2} â€“ 6x + 6 = 0**

**(b) x ^{2} + 6x â€“ 6 = 0**

**(c) x ^{2} â€“ 6x â€“ 6 = 0**

**(d) x ^{2} + 6x + 6 = 0**

**Solution:**

(a) x^{2} â€“ 6x + 6 = 0

Given:

Sum of roots = 6

Product of roots = 6

Thus, the equation is:

x^{2} â€“ 6x + 6 = 0

**Question 14:**

**If \(\alpha\) and \(\beta\) are the roots of the equation 3x ^{2} + 8x + 2 = 0 then \(\left ( \frac{1}{\alpha }+\frac{1}{\beta } \right )\) = ?**

**(a) \(\frac{-3}{8}\)**

**(b) \(\frac{2}{3}\)**

**(c) -4**

**(d) 4**

**Solution:**

It is given that \(\alpha\) and \(\beta\) are the roots of the equation 3x^{2} + 8x + 2 = 0.

Therefore, \(\alpha +\beta =-\frac{8}{3}\) and \(\alpha \beta =\frac{2}{3}\)

\(\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{8}{3}}{\frac{2}{3}}=-4\)

Hence, the correct answer is option C.

**Question 15:**

**The roots of the equation ax ^{2} + bx + c = 0 will be reciprocal of each other if**

**(a) a = b**

**(b) b = c**

**(c) c = a**

**(d) none of these**

**Solution:**

(c) c = a

Let the roots of the equation (ax^{2} + bx + c = 0) be \(\alpha\) and \(\frac{1}{\alpha }\).

Therefore, Product of the roots = \(\alpha\) x \(\frac{1}{\alpha }\) = 1

\(\Rightarrow\) \(\frac{c}{a}=1\)

\(\Rightarrow\) c = a

**Question 16:**

**If the roots of the equation ax ^{2} + bx + c = 0 are equal then c = ?**

**(a) \(\frac{-b}{2a}\)**

**(b) \(\frac{b}{2a}\)**

**(c) \(\frac{-b^{2}}{4a}\)**

**(d) \(\frac{b^{2}}{4a}\)**

**Solution:**

(d) \(\frac{b^{2}}{4a}\)

It is given that the roots of the equation (ax^{2} + bx + c = 0) are equal.

Therefore, (b^{2} â€“ 4ac) = 0

\(\Rightarrow\) b^{2} = 4ac

\(\Rightarrow\) c = \(\frac{b^{2}}{4a}\)

**Question 17:**

**If the equation 9x ^{2} + 6kx + 4 = 0 has equal roots then k = ?**

**(a) 2 or 0**

**(b) -2 or 0**

**(c) 2 or -2**

**(d) 0 only**

**Solution:**

(c) 2 or -2

It is given that the roots of the equation (9x^{2} + 6kx + 4 = 0) are equal.

Therefore, (b^{2} â€“ 4ac) = 0

\(\Rightarrow\) (6k)^{2} â€“ 4 x 9 x 4 = 0

\(\Rightarrow\) 36k^{2} = 144

\(\Rightarrow\) k^{2} = 4

\(\Rightarrow\) k = \(\pm\)2

**Question 18:**

**If the equation x ^{2} + 2(k + 2)x + 9k = 0 has equal roots then k = ?**

**(a) 1 or 4**

**(b) -1 or 4**

**(c) 1 or -4**

**(d) -1 or -4**

**Solution:**

(a) 1 or 4

It is given that the roots of the equation (x^{2} + 2(k + 2)x + 9k = 0) are equal.

Therefore, (b^{2} â€“ 4ac) = 0

\(\Rightarrow\) {2(k + 2)}^{2} â€“ 4 x 1 x 9k = 0

\(\Rightarrow\) 4(k^{2} + 4k + 4) â€“ 36k = 0

\(\Rightarrow\) 4k^{2} + 16k + 16 â€“ 36k = 0

\(\Rightarrow\) 4k^{2} â€“ 20k + 16 = 0

\(\Rightarrow\) k^{2} â€“ 5k + 4 = 0

\(\Rightarrow\) k^{2} â€“ 4k â€“ k + 4 = 0

\(\Rightarrow\) k(k â€“ 4) â€“ (k â€“ 4) = 0

\(\Rightarrow\) (k â€“ 4)(k â€“ 1) = 0

\(\Rightarrow\) k = 4 or k = 1

**Question 19:**

**If the equation 4x ^{2} â€“ 3kx + 1 = 0 has equal roots then k = ?**

**(a) \(\pm \frac{2}{3}\)**

**(b) \(\pm \frac{1}{3}\)**

**(c) \(\pm \frac{3}{4}\)**

**(d) \(\pm \frac{4}{3}\)**

**Solution:**

(d) \(\pm \frac{4}{3}\)

It is given that the roots of the equation (4x^{2} â€“ 3kx + 1 = 0) are equal.

Therefore, (b^{2} â€“ 4ac) = 0

\(\Rightarrow\) (3k^{2}) â€“ 4 x 4 x 1 = 0

\(\Rightarrow\) 9k^{2} = 16

\(\Rightarrow\) k^{2} = \(\frac{16}{9}\)

\(\Rightarrow\) k = \(\pm \frac{4}{3}\)

**Question 20:**

**The roots of ax ^{2} + bx + c = 0, a \(\neq\) 0 are real and unequal, if (b^{2} â€“ 4ac) is**

**(a) > 0**

**(b) = 0**

**(c) < 0**

**(d) none of these**

**Solution:**

(a) > 0

The roots of the equation are real and unequal when (b^{2} â€“ 4ac) > 0.

**Question 21:**

**In the equation ax ^{2} + bx + c = 0, it is given that D = (b^{2} â€“ 4ac) > 0. Then, the roots of the equation are**

**(a) real and equal**

**(b) real and unequal**

**(c) imaginary**

**(d) none of these**

**Solution:**

We know that when discriminant, D > 0, the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

**Question 22:**

**The roots of the equation 2x ^{2} â€“ 6x + 7 = 0 are**

**(a) real, unequal and rational**

**(b) real, unequal and irrational**

**(c) real and equal**

**(d) imaginary**

**Solution:**

(d) imaginary

Because, D = (b^{2} â€“ 4ac)

= (-6)^{2} â€“ 4 x 2 x 7

= 36 â€“ 56

= -20 < 0

Thus, the roots of the equation are imaginary.

**Question 23:**

**The roots of the equation 2x ^{2} â€“ 6x + 3 = 0 are**

**(a) real, unequal and rational**

**(b) real, unequal and irrational**

**(c) real and equal**

**(d) imaginary**

**Solution:**

(b) real, unequal and irrational

Because, D = (b^{2} â€“ 4ac)

= (-6)^{2} â€“ 4 x 2 x 3

= 36 â€“ 24

= 12

12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational.

**Question 24:**

**If the roots of 5x ^{2} â€“ kx + 1 = 0 are rwal and distinct then**

**(a) \(-2\sqrt{5}\) < k < \(2\sqrt{5}\)**

**(b) k > \(2\sqrt{5}\) only**

**(c) k < \(-2\sqrt{5}\) only**

**(d) either k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)**

**Solution:**

(d) either k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)

It is given that the roots of the equation (5x^{2} â€“ kx + 1 = 0) are real and distinct.

Therefore, (b^{2} â€“ 4ac) > 0

\(\Rightarrow\) (-k)^{2} â€“ 4 x 5 x 1 > 0

\(\Rightarrow\) k^{2} â€“ 20 > 0

\(\Rightarrow\) k^{2}> 20

\(\Rightarrow\) k > \(\sqrt{20}\) or k < \(-\sqrt{20}\)

\(\Rightarrow\) k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)

**Question 25:**

**If the equation x ^{2} + 5kx + 16 = 0 has no real roots then**

**(a) k > \(\frac{8}{5}\)**

**(b) k < \(\frac{-8}{5}\)**

**(c) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)**

**(d) none of these**

**Solution:**

(c) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)

It is given that the equation (x^{2} + 5kx + 16 = 0) has no real roots.

Therefore, (b^{2} â€“ 4ac) < 0

\(\Rightarrow\) (5k)^{2} â€“ 4 x 1 x 16 < 0

\(\Rightarrow\) 25k^{2} â€“ 64 < 0

\(\Rightarrow\) k^{2}< \(\frac{64}{25}\)

\(\Rightarrow\) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)

**Question 26:**

**If the equation x ^{2} â€“ kx + 1 = 0 had no real roots then**

**(a) k < -2**

**(b) k > 2**

**(c) -2 < k < 2**

**(d) none of these**

**Solution:**

(c) -2 < k < 2

It is given that the equation x^{2} â€“ kx + 1 = 0 has no real roots.

Therefore, (b^{2} â€“ 4ac) < 0

\(\Rightarrow\) (-k)^{2} â€“ 4 x 1 x 1 < 0

\(\Rightarrow\) k^{2}< 4

\(\Rightarrow\) -2 < k < 2

**Question 27:**

**For what values of k, the equation kx ^{2} â€“ 6x â€“ 2 = 0 has real roots?**

**(a) k \(\leq \frac{-9}{2}\)**

**(b) k \(\geq \frac{-9}{2}\)**

**(c) k \(\leq\) -2**

**(d) none of these**

**Solution:**

(b) k \(\geq \frac{-9}{2}\)

It is given that the roots of the equation (kx^{2} â€“ 6x â€“ 2 = 0) are real.

Therefore, D \(\geq\) 0

\(\Rightarrow\) (b^{2} â€“ 4ac) \(\geq\) 0

\(\Rightarrow\) (-6)^{2} â€“ 4 x k x (-2) \(\geq\) 0

\(\Rightarrow\) 36 + 8k \(\geq\) 0

\(\Rightarrow\) k \(\geq\) \(\frac{-36}{8}\)

\(\Rightarrow\) k \(\geq\) \(\frac{-9}{2}\)

**Question 28:**

**The sum of a number and its reciprocal is \(2\frac{1}{10}\). The number is**

**(a) \(\frac{5}{4}\) or \(\frac{4}{5}\)**

**(b) \(\frac{4}{3}\) or \(\frac{3}{4}\)**

**(c) \(\frac{5}{6}\) or \(\frac{6}{5}\)**

**(d) \(\frac{1}{6}\) or 6**

**Solution:**

(a) \(\frac{5}{4}\) or \(\frac{4}{5}\)

Let the required number be x.

According to the question:

x + \(\frac{1}{x}\) = \(\frac{41}{20}\)

\(\Rightarrow\) \(\frac{x^{2}+1}{x}=\frac{41}{20}\)

\(\Rightarrow\) 20x^{2} â€“ 41x + 20 = 0

\(\Rightarrow\) 20x^{2} â€“ 25x â€“ 16x + 20 = 0

\(\Rightarrow\) 5x(4x â€“ 5) â€“ 4(4x â€“ 5) = 0

\(\Rightarrow\) (4x â€“ 5)(5x â€“ 4) = 0

\(\Rightarrow\) x = \(\frac{5}{4}\) or x = \(\frac{4}{5}\)

**Question 29:**

**The perimeter of a rectangle is 82 m and its area is 400 m ^{2}. The breadth of the rectangle is**

**(a) 25 m**

**(b) 20 m**

**(c) 16 m**

**(d) 9 m**

**Solution:**

(c) 16 m

Let the length and breadth of the rectangle be l and b.

Perimeter of the rectangle = 82 m

\(\Rightarrow\) 2 x (l + b) = 82

\(\Rightarrow\) l + b = 41

\(\Rightarrow\) l = (41 â€“ b) – – – – – – – (i)

Area of the rectangle = 400 m^{2}

\(\Rightarrow\) l x b = 400 m^{2}

\(\Rightarrow\) (41 â€“ b)b = 400 (using (i))

\(\Rightarrow\) 41b â€“ b^{2} = 400

\(\Rightarrow\) b^{2} â€“ 41b + 400 = 0

\(\Rightarrow\) b^{2} â€“ 25b â€“ 16b + 400 = 0

\(\Rightarrow\) b(b â€“ 25) â€“ 16(b â€“ 25) = 0

\(\Rightarrow\) (b â€“ 25)(b â€“ 16) = 0

\(\Rightarrow\) b = 25 or b = 16

If b = 25, we have:

l = 41 â€“ 25 = 16

Since, l cannot be less than b,

Therefore, b = 16 m

**Question 30:**

**The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m ^{2}. The breadth of the field is**

**(a) 20 m**

**(b) 30 m**

**(c) 12 m**

**(d) 16 m**

**Solution:**

Let the breadth of the rectangular filed be x m.

Therefore, Length of the rectangular filed = (x + 8) m

Area of the rectangular filed = 240 m^{2} (Given)

Therefore, \((x+8)\times x=240\) (Area = Length x Breadth)

\(\Rightarrow\) x^{2} + 8x â€“ 240 = 0

\(\Rightarrow\) x^{2} + 20x â€“ 12x â€“ 240 = 0

\(\Rightarrow\) x(x + 20) â€“ 12(x + 20) = 0

\(\Rightarrow\) (x + 20)(x â€“ 12) = 0

\(\Rightarrow\) x + 20 = 0 or x -12 = 0

\(\Rightarrow\) x = -20 or x = 12

Therefore, x = 12 (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

**Question 31:**

**The roots of the quadratic equation 2x ^{2} â€“ x â€“ 6 = 0 are**

**(a) -2, \(\frac{3}{2}\)**

**(b) 2, \(\frac{-3}{2}\)**

**(c) -2, \(\frac{-3}{2}\)**

**(d) 2, \(\frac{3}{2}\)**

**Solution:**

The given quadratic equation is 2x^{2} â€“ x â€“ 6 = 0

2x^{2} â€“ x â€“ 6 = 0

\(\Rightarrow\) 2x^{2} â€“ 4x + 3x â€“ 6 = 0

\(\Rightarrow\) 2x(x â€“ 2) + 3(x â€“ 2) = 0

\(\Rightarrow\) (x â€“ 2)(2x + 3) = 0

\(\Rightarrow\) x â€“ 2 = 0 or 2x + 3 = 0

\(\Rightarrow\) x = 2 or x = \(\frac{-3}{2}\)

Thus, the roots of the given equation are 2 and \(\frac{-3}{2}\).

Hence, the correct answer is option B.

**Question 32:**

**The sum of two natural numbers is 8 and their product is 15. Find the numbers.**

**Solution:**

Let the required numbers be x and (8 â€“ x).

It is given that the product of the two numbers is 15.

Therefore, x(8 â€“ x) = 15

\(\Rightarrow\) 8x â€“ x^{2} = 15

\(\Rightarrow\) x^{2} â€“ 8x + 15 = 0

\(\Rightarrow\) x^{2} â€“ 5x â€“ 3x + 15 = 0

\(\Rightarrow\) x(x â€“ 5) â€“ 3(x â€“ 5) = 0

\(\Rightarrow\)(x â€“ 5)(x â€“ 3) = 0

\(\Rightarrow\) x â€“ 5 = 0 or x â€“ 3 = 0

\(\Rightarrow\) x = 5 or x = 3

Hence, the required numbers are 3 and 5.

**Question 33:**

**Show that x = -3 is a solution of x ^{2} + 6x + 9 = 0.**

**Solution:**

The given equation is x^{2} + 6x + 9 = 0.

Putting x = -3 in the given equation, we get

L.H.S. = (-3)^{2} + 6 x (-3) + 9 = 9 â€“ 18 + 9 = 0 = R.H.S.

Therefore, x = -3 is a solution of the given equation.

**Question 34:**

**Show that x = -2 is a solution of 3x ^{2} + 13x + 14 = 0.**

**Solution:**

The given equation is 3x^{2} + 13x + 14 = 0.

Putting x = -2 in the given equation, we get

L.H.S. = 3 x (-2)^{2} + 13 x (-2) + 14 = 12 â€“ 26 + 14 = 0 = R.H.S.

Therefore, x = -2 is a solution of the given equation.

**Question 35:**

**If x = \(\frac{-1}{2}\) is a solution of the quadratic equation 3x ^{2} + 2kx â€“ 3 = 0, find the value of k.**

**Solution:**

It is given that x = \(\frac{-1}{2}\) is a solution of the quadratic equation 3x^{2} + 2kx â€“ 3 = 0.

Therefore, \(3\times \left ( \frac{-1}{2} \right )^{2}+2k\times \left ( \frac{-1}{2} \right )-3=0\)

\(\Rightarrow\) \(\frac{3}{4}-k-3=0\)

\(\Rightarrow\) k = \(\frac{3-12}{4}=-\frac{9}{4}\)

Hence, the value of k is \(-\frac{9}{4}\).

**Question 36:**

**Find the roots of the quadratic equation 2x ^{2} â€“ x â€“ 6 = 0.**

**Solution:**

The given quadratic equation is 2x^{2} â€“ x â€“ 6 = 0.

2x^{2} â€“ x â€“ 6 = 0

\(\Rightarrow\) 2x^{2} â€“ 4x + 3x â€“ 6 = 0

\(\Rightarrow\) 2x(x â€“ 2) + 3(x â€“ 2) = 0

\(\Rightarrow\) (x â€“ 2)(2x + 3) = 0

\(\Rightarrow\) x â€“ 2 = 0 or 2x + 3 = 0

\(\Rightarrow\) x = 2 or x = \(-\frac{3}{2}\)

Hence, the roots of the given equation are 2 and \(-\frac{3}{2}\).

**Question 37:**

**Find the solution of the quadratic equation \(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\).**

**Solution:**

The given quadratic equation is \(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\)

\(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\)

\(\Rightarrow\) \(3\sqrt{3}x^{2}+9x+x+\sqrt{3}=0\)

\(\Rightarrow\) \(3\sqrt{3}x(x+\sqrt{3})+1(x+\sqrt{3})=0\)

\(\Rightarrow\) \((x+\sqrt{3})(3\sqrt{3}x+1)=0\)

\(\Rightarrow\) \(x+\sqrt{3}=0\) or \(3\sqrt{3}x+1=0\)

\(\Rightarrow\) x = \(-\sqrt{3}\) or x = \(-\frac{1}{3\sqrt{3}}=-\frac{\sqrt{3}}{9}\)

Hence, \(-\sqrt{3}\) and \(-\frac{\sqrt{3}}{9}\) are the solutions of the given equation.

**Question 38:**

**If the roots of the quadratic equation 2x ^{2} + 8x + k = 0 are equal then find the value of k.**

**Solution:**

It is given that the roots of the quadratic equation 2x^{2} + 8k + k = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) 8^{2} â€“ 4 x 2 x k = 0

\(\Rightarrow\) 64 â€“ 8k = 0

\(\Rightarrow\) k = 8

Hence, the value of k is 8.

**Question 39:**

**If the quadratic equation \(px^{2}-2\sqrt{5}px+15=0\) has two equal roots then find the value of p.**

**Solution:**

It is given that quadratic equation \(px^{2}-2\sqrt{5}px+15=0\) has two equal roots.

Therefore, D = 0

\(\Rightarrow\) \((-2\sqrt{5}p)^{2}-4\times p\times 15=0\)

\(\Rightarrow\) 20p^{2} â€“ 60p = 0

\(\Rightarrow\) 20p(p â€“ 3) = 0

\(\Rightarrow\) p = 0 or p â€“ 3 = 0

\(\Rightarrow\) p = 0 or p = 3

For p = 0, we get 15 = 0, which is not true.

Therefore, p \(\neq\) 0

Hence, the value of p is 3.

**Question 40:**

**If 1 is a root of the equation ay ^{2} + ay + 3 = 0 and y^{2} + y + b = 0 then find the value of ab.**

**Solution:**

It is given that y = 1 is a root of the equation ay^{2} + ay + 3 = 0.

Therefore, a x (1)^{2} + a x 1 + 3 = 0

\(\Rightarrow\) a + a + 3 = 0

\(\Rightarrow\) 2a + 3 = 0

\(\Rightarrow\) a = \(-\frac{3}{2}\)

Also, y = 1 is a root of the equation y^{2} + y + b = 0.

Therefore, (1)^{2} + 1 + b = 0

\(\Rightarrow\) 1 + 1 + b = 0

\(\Rightarrow\) b + 2 = 0

\(\Rightarrow\) b = -2

Therefore, ab = \(\left ( -\frac{3}{2} \right )\times (-2)=3\)

Hence, the value of ab is 3.

**Question 41:**

**If one zero of the polynomial x ^{2} â€“ 4x + 1 is \((2+\sqrt{3})\), write the other zero.**

**Solution:**

Let the other zero of the given polynomial be \(\alpha\).

Now,

Sum of the zeroes of the given polynomial = \(\frac{-(-4)}{1}=4\)

Therefore, \(\alpha +(2+\sqrt{3})=4\)

\(\Rightarrow\) \(\alpha =4-2-\sqrt{3}=2-\sqrt{3}\)

Hence, the other zero of the given polynomial is \((2-\sqrt{3})\).

**Question 42:**

**If one root of the quadratic equation 3x ^{2} â€“ 10x + k = 0 is reciprocal of the other, find the value of k.**

**Solution:**

Let \(\alpha\) and \(\beta\) be the roots of the equation 3x^{2} â€“ 10x + k = 0.

Therefore, \(\alpha =\frac{1}{\beta }\) (Given)

\(\Rightarrow\) \(\alpha \beta =1\)

\(\Rightarrow\) \(\frac{k}{3}=1\) (Product of the roots = \(\frac{c}{a}\))

\(\Rightarrow\) k = 3

Hence, the value of k is 3.

**Question 43:**

**If the roots of the quadratic equation px(x â€“ 2) + 6 = 0 are equal, find the value of p.**

**Solution:**

It is given that the roots of the quadratic equation px^{2} â€“ 2px + 6 = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) (-2p)^{2} â€“ 4 x p x 6 = 0

\(\Rightarrow\) 4p^{2} â€“ 24p = 0

\(\Rightarrow\) 4p(p â€“ 6) = 0

\(\Rightarrow\) p = 0 or p â€“ 6 = 0

\(\Rightarrow\) p = 0 or p = 6

For p = 0, we get 6 = 0, which is not true.

Therefore, p \(\neq\) 0

Hence, the value of p is 6.

**Question 44:**

**Find the values of k so that the quadratic equation x ^{2} â€“ 4kx + k = 0 has equal roots.**

**Solution:**

It is given that the quadratic equation x^{2} â€“ 4kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-4k)^{2} â€“ 4 x 1 x k = 0

\(\Rightarrow\) 16k^{2} â€“ 4k = 0

\(\Rightarrow\) 4k(4k â€“ 1) = 0

\(\Rightarrow\) k = 0 or 4k â€“ 1 = 0

\(\Rightarrow\) k = 0 or k = \(\frac{1}{4}\)

Hence, 0 and \(\frac{1}{4}\) are the required values of k.

**Question 45:**

**Find the values of k for which the quadratic equation 9x ^{2} â€“ 3kx + k = 0 has equal roots.**

**Solution:**

It is given that the quadratic equation 9x^{2} â€“ 3kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-3k)^{2} â€“ 4 x 9 x k = 0

\(\Rightarrow\) 9k^{2} â€“ 36k = 0

\(\Rightarrow\) 9k(k â€“ 4) = 0

\(\Rightarrow\) k = 0 or k â€“ 4 = 0

\(\Rightarrow\) k = 0 or k = 4

Hence, 0 and 4 are the required values of k.

**Question 46:**

**Solve: \(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\).**

**Solution:**

\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)

\(\Rightarrow\) \(x^{2}-\sqrt{3}x-x+\sqrt{3}=0\)

\(\Rightarrow\) \(x(x-\sqrt{3})-1(x-\sqrt{3})=0\)

\(\Rightarrow\) \((x-\sqrt{3})(x-1)=0\)

\(\Rightarrow\) \(x-\sqrt{3}=0\) or x â€“ 1 = 0

\(\Rightarrow\) x = \(\sqrt{3}\) or x = 1

Hence, 1 and \(\sqrt{3}\) are the roots of the given equation.

**Question 47:**

**Solve: 2x ^{2} + ax â€“ a^{2} = 0.**

**Solution:**

2x^{2} + ax â€“ a^{2} = 0

\(\Rightarrow\) 2x^{2} + 2ax â€“ ax â€“ a^{2} = 0

\(\Rightarrow\) 2x(x + a) â€“ a(x + a) = 0

\(\Rightarrow\) (x + a)(2x â€“ a) = 0

\(\Rightarrow\) x + a = 0 or 2x â€“ a = 0

\(\Rightarrow\) x = -a or x = \(\frac{a}{2}\)

Hence, -a and \(\frac{a}{2}\) are the roots of the given equation.

**Question 48:**

**Solve: \(3x^{2}+5\sqrt{5}x-10=0\)**

**Solution:**

\(3x^{2}+5\sqrt{5}x-10=0\)

\(\Rightarrow\) \(3x^{2}+6\sqrt{5}x-\sqrt{5}x-10=0\)

\(\Rightarrow\) \(3x(x+2\sqrt{5})-\sqrt{5}(x+2\sqrt{5})=0\)

\(\Rightarrow\) \((x+2\sqrt{5})(3x-\sqrt{5})=0\)

\(\Rightarrow\) \(x+2\sqrt{5}=0\) or \(3x-\sqrt{5}=0\)

\(\Rightarrow\) x = \(-2\sqrt{5}\) or x = \(\frac{\sqrt{5}}{3}\)

Hence, \(-2\sqrt{5}\) and \(\frac{\sqrt{5}}{3}\) are the roots of the given equation.

**Question 49:**

**Solve: \(\sqrt{3}x^{2}+10x-8\sqrt{3}=0\)**

**Solution:**

\(\sqrt{3}x^{2}+10x-8\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x^{2}+12x-2x-8\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x(x+4\sqrt{3})-2(x+4\sqrt{3})=0\)

\(\Rightarrow\) \((x+4\sqrt{3})(\sqrt{3}x-2)=0\)

\(\Rightarrow\) \(x+4\sqrt{3}=0\) or \(\sqrt{3}x-2=0\)

\(\Rightarrow\) x = \(-4\sqrt{3}\) or x = \(\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\)

Hence, \(-4\sqrt{3}\) and \(\frac{2\sqrt{3}}{3}\) are the roots of the given equation.

**Question 50:**

**Solve: \(\sqrt{3}x^{2}-2\sqrt{2}x-2\sqrt{3}=0\)**

**Solution:**

\(\sqrt{3}x^{2}-2\sqrt{2}x-2\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x^{2}-3\sqrt{2}x+2\sqrt{2}x-2\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})\)

\(\Rightarrow\) \((x-\sqrt{6})(\sqrt{3}x+\sqrt{2})=0\)

\(\Rightarrow\) \(x-\sqrt{6}=0\) or \(\sqrt{3}x+\sqrt{2}=0\)

\(\Rightarrow\) x = \(\sqrt{6}\) or x = \(-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}\)

Hence, \(\sqrt{6}\) and \(-\frac{\sqrt{6}}{3}\) are the roots of the given equation.

**Question 51:**

**Solve: \(4\sqrt{3}x^{2}+5x-2\sqrt{3}=0\)**

**Solution:**

\(4\sqrt{3}x^{2}+5x-2\sqrt{3}=0\)

\(\Rightarrow\) \(4\sqrt{3}x^{2}+8x-3x-2\sqrt{3}=0\)

\(\Rightarrow\) \(4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)=0\)

\(\Rightarrow\) \((\sqrt{3}x+2)(4x-\sqrt{3})=0\)

\(\Rightarrow\) \(\sqrt{3}x+2=0\) or \(4x-\sqrt{3}=0\)

\(\Rightarrow\) x = \(-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\) or x = \(\frac{\sqrt{3}}{4}\)

Hence, \(-\frac{2\sqrt{3}}{3}\) and \(\frac{\sqrt{3}}{4}\) are the roots of the given equation.

**Question 52:**

**Solve: 4x ^{2} + 4bx â€“ (a^{2} â€“ b^{2}) = 0**

**Solution:**

4x^{2} + 4bx â€“ (a^{2} â€“ b^{2}) = 0

\(\Rightarrow\) 4x^{2} + 4bx â€“ (a â€“ b)(a + b) = 0

\(\Rightarrow\) 4x^{2} + 2[(a + b) â€“ (a â€“ b)]x â€“ (a â€“ b)(a + b) = 0

\(\Rightarrow\) 4x^{2} + 2(a + b)x â€“ 2(a â€“ b)x â€“ (a â€“ b)(a + b) = 0

\(\Rightarrow\) 2x[2x + (a + b)] â€“ (a â€“ b)[2x + (a + b)] = 0

\(\Rightarrow\) [2x + (a + b)][2x â€“ (a â€“ b)] = 0

\(\Rightarrow\) 2x + (a + b) = 0 or 2x â€“ (a â€“ b) = 0

\(\Rightarrow\) x = \(-\frac{a+b}{2}\) or x = \(\frac{a-b}{2}\)

Hence, \(-\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are the roots of the given equation.

**Question 53:**

**Solve: x ^{2} + 5x â€“ (a^{2} + a â€“ 6) = 0**

**Solution:**

x^{2} + 5x â€“ (a^{2} + a â€“ 6) = 0

\(\Rightarrow\) x^{2} + 5x â€“ (a + 3)(a â€“ 2) = 0

\(\Rightarrow\) x^{2} + [(a + 3) â€“ (a â€“ 2)]x â€“ (a + 3)(a â€“ 2) = 0

\(\Rightarrow\) x^{2} + (a + 3)x â€“ (a â€“ 2)x â€“ (a + 3)(a â€“ 2) = 0

\(\Rightarrow\) x[x + (a + 3)] â€“ (a â€“ 2)[x + (a + 3)] = 0

\(\Rightarrow\) [x + (a + 3)][x â€“ (a â€“ 2)] = 0

\(\Rightarrow\) x + (a + 3) = 0 or x â€“ (a â€“ 2) = 0

\(\Rightarrow\) x = -(a + 3) or x = (a â€“ 2)

Hence, -(a + 3) and (a â€“ 2) are the roots of the given equation.

**Question 54:**

**x ^{2} + 6x â€“ (a^{2} + 2a â€“ 8) = 0**

**Solution:**

x^{2} + 6x â€“ (a^{2} + 2a â€“ 8) = 0

\(\Rightarrow\) x^{2} + 6x â€“ (a^{2} + 2a â€“ 8) = 0

\(\Rightarrow\) x^{2} + [(a + 4) â€“ (a â€“ 2)]x â€“ (a + 4)(a â€“ 2) = 0

\(\Rightarrow\) x^{2} + (a + 4)x â€“ (a â€“ 2)x â€“ (a + 4)(a â€“ 2) = 0

\(\Rightarrow\) x[x + (a + 4) â€“ (a â€“ 2)[x + (a + 4)] = 0

\(\Rightarrow\) [x + (a + 4)][x â€“ (a â€“ 2)] = 0

\(\Rightarrow\) x + (a + 4) = 0 or x â€“ (a â€“ 2) = 0

\(\Rightarrow\) x = -(a + 4) or x = (a â€“ 2)

Hence, -(a + 4) and (a â€“ 2) are the roots of the given equation.

**Question 55:**

**x ^{2} â€“ 4ax + 4a^{2} â€“ b^{2} = 0**

**Solution:**

x^{2} â€“ 4ax + 4a^{2} â€“ b^{2} = 0

\(\Rightarrow\) x^{2} â€“ 4ax + (2a + b)(2a â€“ b) = 0

\(\Rightarrow\) x^{2} â€“ [(2a + b) (2a â€“ b)]x + (2a + b)(2a â€“ b) = 0

\(\Rightarrow\) x^{2} â€“ (2a + b)x â€“ (2a â€“ b)x + (2a + b)(2a â€“ b) = 0

\(\Rightarrow\) x[x â€“ (2a + b)] â€“ (2a â€“ b)[x â€“ (2a + b)] = 0

\(\Rightarrow\) [x â€“ (2a + b)][x â€“ (2a â€“ b) = 0

\(\Rightarrow\) x â€“ (2a + b) = 0 or x â€“ (2a â€“ b) = 0

\(\Rightarrow\) x = (2a + b) or x = (2a â€“ b)

Hence, (2a + b) and (2a â€“ b) are the roots of the given equation.

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