**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} – 5x = (x – 1)^{2}

\(\Rightarrow\)^{2} – 5x = x^{2} – 2x + 1

\(\Rightarrow\)^{2} – 5x – x^{2} + 2x – 1 = 0

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

**Question 2:**

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

**(a) (x ^{2} + 1) = (2 – x)^{2} + 3**

**(b) x ^{3} – x^{2} = (x – 1)^{3}**

**(c) 2x ^{2} + 3 = (5 + x)(2x – 3)**

**(d) none of these**

**Solution:**

(b) x^{3} – x^{2} = (x – 1)^{3}

Because, x^{3} – x^{2} = (x – 1)^{3}

\(\Rightarrow\)^{3} – x^{2} = x^{3} – 3x^{2} + 3x – 1

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

**Question 3:**

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

**(a) 3x – x ^{2} = x^{2} + 5**

**(b) (x + 2) ^{2} = 2(x^{2} – 5)**

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

**(d) (x – 1) ^{2} = 3x^{2} + x – 2**

**Solution:**

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

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

\(\Rightarrow\)

\(\Rightarrow\)

**Question 4:**

**If x = 3 is a solution of the equation 3x ^{2} + (k – 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 – 1)x + 9 = 0; therefore, we have:

3(3)^{2} + (k – 1) x 3 + 9 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**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

\(\Rightarrow\)

\(\Rightarrow\)

**Question 6:**

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

**(a) 2**

**(b) -2**

**(c) 6**

**(d) -6**

**Solution:**

(c) 6

Sum of the roots of the equation x^{2} – 6x + 2 = 0 is

\(\alpha +\beta =\frac{-b}{a}=\frac{-(-6)}{1}=6\)

**Question 7:**

**If the product of the roots of the equation x ^{2} – 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} – 3x + k = 10 is -2.

The equation can be rewritten as:

x^{2} – 3x + (k – 10) = 0

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**Question 8:**

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

**(a) 7 : 12**

**(b) 7 : 18**

**(c) 3 : 2**

**(d) 2 : 3**

**Solution:**

(c) 2 : 3

Given:

7x^{2} – 12 x + 18 = 0

Therefore, \(\alpha +\beta =\frac{12}{7}\)

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} – 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} – 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 = \(\frac{c}{a}\)

\(\Rightarrow\)

\(\Rightarrow\)

**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 \(\alpha\)

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**Question 12:**

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

**(a) x ^{2} – 3x + 10 = 0**

**(b) x ^{2} – 3x – 10 = 0**

**(c) x ^{2} + 3x – 10 = 0**

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

**Solution:**

(b) x^{2} – 3x – 10 = 0

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

Then, the equation is:

x^{2} – (5 – 2)x + 5 x (-2) = 0

\(\Rightarrow\)^{2} – 3x – 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} – 6x + 6 = 0**

**(b) x ^{2} + 6x – 6 = 0**

**(c) x ^{2} – 6x – 6 = 0**

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

**Solution:**

(a) x^{2} – 6x + 6 = 0

Given:

Sum of roots = 6

Product of roots = 6

Thus, the equation is:

x^{2} – 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 \(\alpha\)^{2} + 8x + 2 = 0.

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

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

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 \(\alpha\)

Therefore, Product of the roots = \(\alpha\)

\(\Rightarrow\)

\(\Rightarrow\)

**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} – 4ac) = 0

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

\(\Rightarrow\)

**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} – 4ac) = 0

\(\Rightarrow\)^{2} – 4 x 9 x 4 = 0

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

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

\(\Rightarrow\)

**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} – 4ac) = 0

\(\Rightarrow\)^{2} – 4 x 1 x 9k = 0

\(\Rightarrow\)^{2} + 4k + 4) – 36k = 0

\(\Rightarrow\)^{2} + 16k + 16 – 36k = 0

\(\Rightarrow\)^{2} – 20k + 16 = 0

\(\Rightarrow\)^{2} – 5k + 4 = 0

\(\Rightarrow\)^{2} – 4k – k + 4 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**Question 19:**

**If the equation 4x ^{2} – 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} – 3kx + 1 = 0) are equal.

Therefore, (b^{2} – 4ac) = 0

\(\Rightarrow\)^{2}) – 4 x 4 x 1 = 0

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

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

\(\Rightarrow\)

**Question 20:**

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

**Question 21:**

**In the equation ax ^{2} + bx + c = 0, it is given that D = (b^{2} – 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} – 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} – 4ac)

= (-6)^{2} – 4 x 2 x 7

= 36 – 56

= -20 < 0

Thus, the roots of the equation are imaginary.

**Question 23:**

**The roots of the equation 2x ^{2} – 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} – 4ac)

= (-6)^{2} – 4 x 2 x 3

= 36 – 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} – 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 > \(2\sqrt{5}\)

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

Therefore, (b^{2} – 4ac) > 0

\(\Rightarrow\)^{2} – 4 x 5 x 1 > 0

\(\Rightarrow\)^{2} – 20 > 0

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

\(\Rightarrow\)

\(\Rightarrow\)

**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) \(\frac{-8}{5}\)

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

Therefore, (b^{2} – 4ac) < 0

\(\Rightarrow\)^{2} – 4 x 1 x 16 < 0

\(\Rightarrow\)^{2} – 64 < 0

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

\(\Rightarrow\)

**Question 26:**

**If the equation x ^{2} – 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} – kx + 1 = 0 has no real roots.

Therefore, (b^{2} – 4ac) < 0

\(\Rightarrow\)^{2} – 4 x 1 x 1 < 0

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

\(\Rightarrow\)

**Question 27:**

**For what values of k, the equation kx ^{2} – 6x – 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} – 6x – 2 = 0) are real.

Therefore, D \(\geq\)

\(\Rightarrow\)^{2} – 4ac) \(\geq\)

\(\Rightarrow\)^{2} – 4 x k x (-2) \(\geq\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

Let the required number be x.

According to the question:

x + \(\frac{1}{x}\)

\(\Rightarrow\)

\(\Rightarrow\)^{2} – 41x + 20 = 0

\(\Rightarrow\)^{2} – 25x – 16x + 20 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

**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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Area of the rectangle = 400 m^{2}

\(\Rightarrow\)^{2}

\(\Rightarrow\)

\(\Rightarrow\)^{2} = 400

\(\Rightarrow\)^{2} – 41b + 400 = 0

\(\Rightarrow\)^{2} – 25b – 16b + 400 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

If b = 25, we have:

l = 41 – 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, \((x+8)\times x=240\)

\(\Rightarrow\)^{2} + 8x – 240 = 0

\(\Rightarrow\)^{2} + 20x – 12x – 240 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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} – x – 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} – x – 6 = 0

2x^{2} – x – 6 = 0

\(\Rightarrow\)^{2} – 4x + 3x – 6 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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 – x).

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

Therefore, x(8 – x) = 15

\(\Rightarrow\)^{2} = 15

\(\Rightarrow\)^{2} – 8x + 15 = 0

\(\Rightarrow\)^{2} – 5x – 3x + 15 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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 – 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 – 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 – 3 = 0, find the value of k.**

**Solution:**

It is given that x = \(\frac{-1}{2}\)^{2} + 2kx – 3 = 0.

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

\(\Rightarrow\)

\(\Rightarrow\)

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

**Question 36:**

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

**Solution:**

The given quadratic equation is 2x^{2} – x – 6 = 0.

2x^{2} – x – 6 = 0

\(\Rightarrow\)^{2} – 4x + 3x – 6 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(-\sqrt{3}\)

**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

\(\Rightarrow\)^{2} – 4 x 2 x k = 0

\(\Rightarrow\)

\(\Rightarrow\)

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 \(px^{2}-2\sqrt{5}px+15=0\)

Therefore, D = 0

\(\Rightarrow\)

\(\Rightarrow\)^{2} – 60p = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, p \(\neq\)

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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} – 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, \(\alpha +(2+\sqrt{3})=4\)

\(\Rightarrow\)

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

**Question 42:**

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

**Solution:**

Let \(\alpha\)^{2} – 10x + k = 0.

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, the value of k is 3.

**Question 43:**

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

**Solution:**

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

Therefore, D = 0

\(\Rightarrow\)^{2} – 4 x p x 6 = 0

\(\Rightarrow\)^{2} – 24p = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, p \(\neq\)

Hence, the value of p is 6.

**Question 44:**

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

**Solution:**

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

Therefore, D = 0

\(\Rightarrow\)^{2} – 4 x 1 x k = 0

\(\Rightarrow\)^{2} – 4k = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, 0 and \(\frac{1}{4}\)

**Question 45:**

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

**Solution:**

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

Therefore, D = 0

\(\Rightarrow\)^{2} – 4 x 9 x k = 0

\(\Rightarrow\)^{2} – 36k = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

**Question 46:**

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

**Solution:**

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, 1 and \(\sqrt{3}\)

**Question 47:**

**Solve: 2x ^{2} + ax – a^{2} = 0.**

**Solution:**

2x^{2} + ax – a^{2} = 0

\(\Rightarrow\)^{2} + 2ax – ax – a^{2} = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, -a and \(\frac{a}{2}\)

**Question 48:**

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

**Solution:**

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(-2\sqrt{5}\)

**Question 49:**

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

**Solution:**

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(-4\sqrt{3}\)

**Question 50:**

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

**Solution:**

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(\sqrt{6}\)

**Question 51:**

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

**Solution:**

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(-\frac{2\sqrt{3}}{3}\)

**Question 52:**

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

**Solution:**

4x^{2} + 4bx – (a^{2} – b^{2}) = 0

\(\Rightarrow\)^{2} + 4bx – (a – b)(a + b) = 0

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

\(\Rightarrow\)^{2} + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, \(-\frac{a+b}{2}\)

**Question 53:**

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

**Solution:**

x^{2} + 5x – (a^{2} + a – 6) = 0

\(\Rightarrow\)^{2} + 5x – (a + 3)(a – 2) = 0

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

\(\Rightarrow\)^{2} + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

**Question 54:**

**x ^{2} + 6x – (a^{2} + 2a – 8) = 0**

**Solution:**

x^{2} + 6x – (a^{2} + 2a – 8) = 0

\(\Rightarrow\)^{2} + 6x – (a^{2} + 2a – 8) = 0

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

\(\Rightarrow\)^{2} + (a + 4)x – (a – 2)x – (a + 4)(a – 2) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

**Question 55:**

**x ^{2} – 4ax + 4a^{2} – b^{2} = 0**

**Solution:**

x^{2} – 4ax + 4a^{2} – b^{2} = 0

\(\Rightarrow\)^{2} – 4ax + (2a + b)(2a – b) = 0

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

\(\Rightarrow\)^{2} – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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