RS Aggarwal Solutions Class 10 Ex 10H

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, 2x2 – 5x = (x – 1)2

\(\Rightarrow\) 2x2 – 5x = x2 – 2x + 1

\(\Rightarrow\) 2x2 – 5x – x2 + 2x – 1 = 0

\(\Rightarrow\) x2 – 3x – 1 = 0, which is a quadratic equation

Question 2:

Which of the following is a quadratic equation?

(a) (x2 + 1) = (2 – x)2 + 3

(b) x3 – x2 = (x – 1)3

(c) 2x2 + 3 = (5 + x)(2x – 3)

(d) none of these

Solution:

(b) x3 – x2 = (x – 1)3

Because, x3 – x2 = (x – 1)3

\(\Rightarrow\) x3 – x2 = x3 – 3x2 + 3x – 1

\(\Rightarrow\) 2x2 – 3x + 1 = 0, which is a quadratic equation

Question 3:

Which of the following is not a quadratic equation?

(a) 3x – x2 = x2 + 5

(b) (x + 2)2 = 2(x2 – 5)

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

(d) (x – 1)2 = 3x2 + x – 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 3x2 + (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 3x2 + (k – 1)x + 9 = 0; therefore, we have:

3(3)2 + (k – 1) x 3 + 9 = 0

\(\Rightarrow\) 27 + 3(k – 1) + 9 = 0

\(\Rightarrow\) 3(k – 1) = -36

\(\Rightarrow\) (k – 1) = -12

\(\Rightarrow\) k = -11

Question 5:

If one root of the equation 2x2 + 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 2x2 + ax + 6 = 0 is 2.

Therefore, 2 x 22 + a x 2 + 6 = 0

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

\(\Rightarrow\) a = -7

Question 6:

The sum of the roots of the equation x2 – 6x + 2 = 0 is

(a) 2

(b) -2

(c) 6

(d) -6

Solution:

(c) 6

Sum of the roots of the equation x2 – 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 x2 – 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 x2 – 3x + k = 10 is -2.

The equation can be rewritten as:

x2 – 3x + (k – 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 7x2 – 12x + 18 = 0 is

(a) 7 : 12

(b) 7 : 18

(c) 3 : 2

(d) 2 : 3

Solution:

(c) 2 : 3

Given:

7x2 – 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 3x2 – 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:

3x2 – 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 5x2 + 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 (5x2 + 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 kx2 + 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:

kx2 + 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) x2 – 3x + 10 = 0

(b) x2 – 3x – 10 = 0

(c) x2 + 3x – 10 = 0

(d) x2 + 3x + 10 = 0

Solution:

(b) x2 – 3x – 10 = 0

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

Then, the equation is:

x2 – (5 – 2)x + 5 x (-2) = 0

\(\Rightarrow\) x2 – 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) x2 – 6x + 6 = 0

(b) x2 + 6x – 6 = 0

(c) x2 – 6x – 6 = 0

(d) x2 + 6x + 6 = 0

Solution:

(a) x2 – 6x + 6 = 0

Given:

Sum of roots = 6

Product of roots = 6

Thus, the equation is:

x2 – 6x + 6 = 0

Question 14:

If \(\alpha\) and \(\beta\) are the roots of the equation 3x2 + 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 3x2 + 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 ax2 + 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 (ax2 + 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 ax2 + 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 (ax2 + bx + c = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) b2 = 4ac

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

Question 17:

If the equation 9x2 + 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 (9x2 + 6kx + 4 = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) (6k)2 – 4 x 9 x 4 = 0

\(\Rightarrow\) 36k2 = 144

\(\Rightarrow\) k2 = 4

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

Question 18:

If the equation x2 + 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 (x2 + 2(k + 2)x + 9k = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) {2(k + 2)}2 – 4 x 1 x 9k = 0

\(\Rightarrow\) 4(k2 + 4k + 4) – 36k = 0

\(\Rightarrow\) 4k2 + 16k + 16 – 36k = 0

\(\Rightarrow\) 4k2 – 20k + 16 = 0

\(\Rightarrow\) k2 – 5k + 4 = 0

\(\Rightarrow\) k2 – 4k – k + 4 = 0

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

\(\Rightarrow\) (k – 4)(k – 1) = 0

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

Question 19:

If the equation 4x2 – 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 (4x2 – 3kx + 1 = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) (3k2) – 4 x 4 x 1 = 0

\(\Rightarrow\) 9k2 = 16

\(\Rightarrow\) k2 = \(\frac{16}{9}\)

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

Question 20:

The roots of ax2 + bx + c = 0, a \(\neq\) 0 are real and unequal, if (b2 – 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 (b2 – 4ac) > 0.

Question 21:

In the equation ax2 + bx + c = 0, it is given that D = (b2 – 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 2x2 – 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 = (b2 – 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 2x2 – 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 = (b2 – 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 5x2 – 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 (5x2 – kx + 1 = 0) are real and distinct.

Therefore, (b2 – 4ac) > 0

\(\Rightarrow\) (-k)2 – 4 x 5 x 1 > 0

\(\Rightarrow\) k2 – 20 > 0

\(\Rightarrow\) k2> 20

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

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

Question 25:

If the equation x2 + 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 (x2 + 5kx + 16 = 0) has no real roots.

Therefore, (b2 – 4ac) < 0

\(\Rightarrow\) (5k)2 – 4 x 1 x 16 < 0

\(\Rightarrow\) 25k2 – 64 < 0

\(\Rightarrow\) k2< \(\frac{64}{25}\)

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

Question 26:

If the equation x2 – 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 x2 – kx + 1 = 0 has no real roots.

Therefore, (b2 – 4ac) < 0

\(\Rightarrow\) (-k)2 – 4 x 1 x 1 < 0

\(\Rightarrow\) k2< 4

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

Question 27:

For what values of k, the equation kx2 – 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 (kx2 – 6x – 2 = 0) are real.

Therefore, D \(\geq\) 0

\(\Rightarrow\) (b2 – 4ac) \(\geq\) 0

\(\Rightarrow\) (-6)2 – 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 + \(\frac{1}{x}\) = \(\frac{41}{20}\)

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

\(\Rightarrow\) 20x2 – 41x + 20 = 0

\(\Rightarrow\) 20x2 – 25x – 16x + 20 = 0

\(\Rightarrow\) 5x(4x – 5) – 4(4x – 5) = 0

\(\Rightarrow\) (4x – 5)(5x – 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 m2. 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 – b) – – – – – – – (i)

Area of the rectangle = 400 m2

\(\Rightarrow\) l x b = 400 m2

\(\Rightarrow\) (41 – b)b = 400 (using (i))

\(\Rightarrow\) 41b – b2 = 400

\(\Rightarrow\) b2 – 41b + 400 = 0

\(\Rightarrow\) b2 – 25b – 16b + 400 = 0

\(\Rightarrow\) b(b – 25) – 16(b – 25) = 0

\(\Rightarrow\) (b – 25)(b – 16) = 0

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

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 m2. 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 m2 (Given)

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

\(\Rightarrow\) x2 + 8x – 240 = 0

\(\Rightarrow\) x2 + 20x – 12x – 240 = 0

\(\Rightarrow\) x(x + 20) – 12(x + 20) = 0

\(\Rightarrow\) (x + 20)(x – 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 2x2 – 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 2x2 – x – 6 = 0

2x2 – x – 6 = 0

\(\Rightarrow\) 2x2 – 4x + 3x – 6 = 0

\(\Rightarrow\) 2x(x – 2) + 3(x – 2) = 0

\(\Rightarrow\) (x – 2)(2x + 3) = 0

\(\Rightarrow\) x – 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 – x).

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

Therefore, x(8 – x) = 15

\(\Rightarrow\) 8x – x2 = 15

\(\Rightarrow\) x2 – 8x + 15 = 0

\(\Rightarrow\) x2 – 5x – 3x + 15 = 0

\(\Rightarrow\) x(x – 5) – 3(x – 5) = 0

\(\Rightarrow\)(x – 5)(x – 3) = 0

\(\Rightarrow\) x – 5 = 0 or x – 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 x2 + 6x + 9 = 0.

Solution:

The given equation is x2 + 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 3x2 + 13x + 14 = 0.

Solution:

The given equation is 3x2 + 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 3x2 + 2kx – 3 = 0, find the value of k.

Solution:

It is given that x = \(\frac{-1}{2}\) is a solution of the quadratic equation 3x2 + 2kx – 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 2x2 – x – 6 = 0.

Solution:

The given quadratic equation is 2x2 – x – 6 = 0.

2x2 – x – 6 = 0

\(\Rightarrow\) 2x2 – 4x + 3x – 6 = 0

\(\Rightarrow\) 2x(x – 2) + 3(x – 2) = 0

\(\Rightarrow\) (x – 2)(2x + 3) = 0

\(\Rightarrow\) x – 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 2x2 + 8x + k = 0 are equal then find the value of k.

Solution:

It is given that the roots of the quadratic equation 2x2 + 8k + k = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) 82 – 4 x 2 x k = 0

\(\Rightarrow\) 64 – 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

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

\(\Rightarrow\) 20p2 – 60p = 0

\(\Rightarrow\) 20p(p – 3) = 0

\(\Rightarrow\) p = 0 or p – 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 ay2 + ay + 3 = 0 and y2 + y + b = 0 then find the value of ab.

Solution:

It is given that y = 1 is a root of the equation ay2 + 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 y2 + 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 x2 – 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 3x2 – 10x + k = 0 is reciprocal of the other, find the value of k.

Solution:

Let \(\alpha\) and \(\beta\) be the roots of the equation 3x2 – 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 – 2) + 6 = 0 are equal, find the value of p.

Solution:

It is given that the roots of the quadratic equation px2 – 2px + 6 = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) (-2p)2 – 4 x p x 6 = 0

\(\Rightarrow\) 4p2 – 24p = 0

\(\Rightarrow\) 4p(p – 6) = 0

\(\Rightarrow\) p = 0 or p – 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 x2 – 4kx + k = 0 has equal roots.

Solution:

It is given that the quadratic equation x2 – 4kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-4k)2 – 4 x 1 x k = 0

\(\Rightarrow\) 16k2 – 4k = 0

\(\Rightarrow\) 4k(4k – 1) = 0

\(\Rightarrow\) k = 0 or 4k – 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 9x2 – 3kx + k = 0 has equal roots.

Solution:

It is given that the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-3k)2 – 4 x 9 x k = 0

\(\Rightarrow\) 9k2 – 36k = 0

\(\Rightarrow\) 9k(k – 4) = 0

\(\Rightarrow\) k = 0 or k – 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 – 1 = 0

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

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

Question 47:

Solve: 2x2 + ax – a2 = 0.

Solution:

2x2 + ax – a2 = 0

\(\Rightarrow\) 2x2 + 2ax – ax – a2 = 0

\(\Rightarrow\) 2x(x + a) – a(x + a) = 0

\(\Rightarrow\) (x + a)(2x – a) = 0

\(\Rightarrow\) x + a = 0 or 2x – 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: 4x2 + 4bx – (a2 – b2) = 0

Solution:

4x2 + 4bx – (a2 – b2) = 0

\(\Rightarrow\) 4x2 + 4bx – (a – b)(a + b) = 0

\(\Rightarrow\) 4x2 + 2[(a + b) – (a – b)]x – (a – b)(a + b) = 0

\(\Rightarrow\) 4x2 + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

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

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

\(\Rightarrow\) 2x + (a + b) = 0 or 2x – (a – 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: x2 + 5x – (a2 + a – 6) = 0

Solution:

x2 + 5x – (a2 + a – 6) = 0

\(\Rightarrow\) x2 + 5x – (a + 3)(a – 2) = 0

\(\Rightarrow\) x2 + [(a + 3) – (a – 2)]x – (a + 3)(a – 2) = 0

\(\Rightarrow\) x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

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

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

\(\Rightarrow\) x + (a + 3) = 0 or x – (a – 2) = 0

\(\Rightarrow\) x = -(a + 3) or x = (a – 2)

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

Question 54:

x2 + 6x – (a2 + 2a – 8) = 0

Solution:

x2 + 6x – (a2 + 2a – 8) = 0

\(\Rightarrow\) x2 + 6x – (a2 + 2a – 8) = 0

\(\Rightarrow\) x2 + [(a + 4) – (a – 2)]x – (a + 4)(a – 2) = 0

\(\Rightarrow\) x2 + (a + 4)x – (a – 2)x – (a + 4)(a – 2) = 0

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

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

\(\Rightarrow\) x + (a + 4) = 0 or x – (a – 2) = 0

\(\Rightarrow\) x = -(a + 4) or x = (a – 2)

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

Question 55:

x2 – 4ax + 4a2 – b2 = 0

Solution:

x2 – 4ax + 4a2 – b2 = 0

\(\Rightarrow\) x2 – 4ax + (2a + b)(2a – b) = 0

\(\Rightarrow\) x2 – [(2a + b) (2a – b)]x + (2a + b)(2a – b) = 0

\(\Rightarrow\) x2 – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0

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

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

\(\Rightarrow\) x – (2a + b) = 0 or x – (2a – b) = 0

\(\Rightarrow\) x = (2a + b) or x = (2a – b)

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


Practise This Question

Mendeleev predicted the existence of some elements that had not been discovered at that time and left some gaps in his periodic table. What are the names that he gave to the two undiscovered elements, gallium and germanium, respectively?