# RS Aggarwal Solutions Class 10 Ex 2B

Q.1: If one zero of the polynomial $x^{2} + x – 4x + 1$ is $(2+ \sqrt{3})$, write the other zero.

Sol:

Let the other zeros of $x^{2} + x – 4x + 1$ be a.

By using the relationship between the zeros of the quadratic polynomial

We have, Sum of zeros = $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$

$Therefore, 2 + \sqrt{3} + a = \frac{-(-4)}{1}$

$\Rightarrow a = 2 – \sqrt{3}$

Hence, the other zeros of $x^{2} + x – 4x + 1$ is $2 – \sqrt{3}$

Q.2: Find the zeros of the polynomial $x^{2} + x – p(p + 1)$.

Sol:

f(x) = $x^{2} + x – p(p + 1)$

By adding and subtracting px, we have

f(x) = $x^{2} + px + x – px – p(p+1)$

$= x^{2} + (p+1) x – px – p(p+1)$

$= x \left [ x + (p+1) \right ] – p \left [ x + (p+1) \right ]$

= $= \left [ x + (p+1) \right ] (x – p)$

$f(x) = 0$

$\Rightarrow \left [ x + (p+1) \right ] (x – p) = 0$

$\Rightarrow \left [ x + (p+1) \right ] = 0 \;\; or \;\;(x – p) = 0$

$\Rightarrow x = -(p+1) \;\; or \;\;x = p$

So, the zeros of f(x) are -(p+1) and p.

Q.3: Find the zeros of the polynomial $x^{2} – 3 x – m(m + 3)$.

Sol:

f(x) = $x^{2} – 3 x – m(m + 3)$

By adding and subtracting mx , we have

f(x) = $x^{2} – mx – 3 x + mx – m(m + 3)$

$= x \left [ x – (m+3) \right ](x + m) + m \left [ x – (m + 3) \right ]$

$= \left [ x -(m+3) \right ] (x + m)$

f (x) = 0

$\Rightarrow \left [ x -(m+3) \right ] (x + m) = 0$

$\Rightarrow \left [ x -(m+3) \right ] = 0 \;\; or \;\; (x + m) = 0$

$\Rightarrow x = m + 3 \;\; or \;\; x = -m$

So, the zeros of f(x) are -m and m+3.

Q.4: If $\alpha \;\; and \;\; \beta$ are the zeros of the polynomial such that $\alpha + \beta = 6$ and $\alpha \beta = 4$

Then write the polynomial.

Sol:

If $\alpha \;\; and \;\; \beta$ are the zeros of the polynomial then the quadratic polynomial can be found as

$x^{2} – (\alpha + \beta )x + \alpha \beta$   ….(i)

Substituting the values from the given conditions

$x^{2} – 6x + 4$

Q.5: If one zero of the Quadratic polynomial $kx^{2} + 3x + k \;\; is \;\; 2$, then find the value of k.

Sol:

If x =2 is one zero of the quadratic polynomial $kx^{2} + 3x + k$

Therefore,  it will satisfy the above polynomial.

Now, we have

$k(2)^{2} + 3(2) + k = 0$

$\Rightarrow 4k + 6 + k = 0$

$\Rightarrow 5k + 6 = 0$

$\Rightarrow k = -\frac{6}{5}$

Q.6: If 3 is a zero of the polynomial $2x^{2} + x + k \;\; is \;\; 2$, then find the value of k .

Sol:

Given x=3 is one zero of the polynomial $2x^{2} + x + k$

Therefore it will satisfy the above polynomial.

Now, we have

$2(3)^{2} + 3 + k = 0$

$\Rightarrow 21 + k = 0$

$\Rightarrow k = -21$

Q.7: If -4 is a zero of the polynomial $x^{2} – x – (2k + 2)$ then find the value of k .

Sol:

Given x= -4 is one zero of the polynomial $x^{2} – x – (2k + 2)$

Therefore it will satisfy the above polynomial.

Now, we have

$(-4)^{2} – (-4) – (2k + 2) = 0$

$\Rightarrow 16 + 4 – 2k – 2 = 0$

$\Rightarrow – 2k = -18$

$\Rightarrow k = 9$

Q.8: If 1 is a zero of the polynomial $ax^{2} – 3(a-1)x -1$ then find the value of a.

Sol:

Given x= 1 is one zero of the polynomial $ax^{2} – 3(a-1)x -1$

Therefore it will satisfy the above polynomial.

Now, we have

$a(1)^{2} – 3(a-1)1 -1 = 0$

$\Rightarrow a – 3a + 3 – 1 = 0$

$\Rightarrow – 2a = -2$

$\Rightarrow a = 1$

Q.9: If -2 is a zero of the polynomial $3x^{2} + 4x + 2k$ then find the value of k.

Sol:

Given x= -2 is one zero of the polynomial $3x^{2} + 4x + 2k$

Therefore it will satisfy the above polynomial.

Now, we have

$3(-2)^{2} + 4 (-2) + 2k = 0$

$\Rightarrow 12 – 8 + 2k = 0$

$\Rightarrow k = -2$

Q.10: Write the zeros of the polynomial $x^{2} – x – 6$

Sol:

$f(x) = x^{2} – x – 6$

$= x^{2} – 3x + 2x – 6$

$= x(x – 3) + 2( x- 3)$

$= (x – 3) ( x – 2)$

$f(x) = 0 \\ \Rightarrow (x – 3) ( x – 2) = 0 \\ \Rightarrow (x – 3) = 0 \;\; or \;\; (x-2) = 0 \\ \Rightarrow x = 3 \;\; or \;\; x = -2$

Q.11: If the sum of the zeros of the Quadratic polynomial $kx^{2} – 3x + 5$ is 1, write the value of k.

Sol:

By using the relationship between the zeros of the polynomial,

We have

Sum of zeros = $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$

$\Rightarrow 1 = \frac{- (-3)}{k}$

$\Rightarrow k = 3$

Q.12: If the product of the zeros of the Quadratic polynomial $x^{2} – 4x + k$ is 3 then write the value of k.

Sol:

By using the relationship between the zeros of the quadratic polynomial

We have

Product of zeros = $\frac{Constant \; term}{ Coefficient \; of \; x^{2}}$

$\Rightarrow 3 = \frac{k}{1}$

$\Rightarrow k = 3$

Q.13: If (x+a) is a factor of $2x^{2} + 2ax + 5x + 10$, find the value of  a.

Sol:

Given (x+a) is a factor of $2x^{2} + 2ax + 5x + 10$

We have

x + a = 0

x = -a

Since (x + a) is a factor of $2x^{2} + 2ax + 5x + 10$

Hence, it will satisfy the above polynomial

$Therefore, 2(-a)^{2} + 2a (-a) + 5(-a) + 10 = 0$

$\Rightarrow -5a + 10 = 0$

$\Rightarrow a = 2$

Q.14: If (a – b) , a and ( a + b) area zeros of the polynomial $2x^{3} – 6x^{2} + 5x – 7$ then write the value of a.

Sol:

By using the relationship between the zeros of the cubic polynomial

We have,

Sum of zeros = $\frac{- (Coefficient \; of \; x^{2} )}{Coefficient \; of \; x^{3}}$

$\Rightarrow a – b + a + a + b = \frac{-(-6)}{2}$

$\Rightarrow 3a = 3$

$\Rightarrow a = 1$

Q.15: If $x^{3} + x^{2} – ax + b$ is divisible by $x^{2} – x$, write the value of x.

Sol:

Equating $x^{2} – x$ to find the zeros, we get

$x (x – 1) = 0$

$\Rightarrow x = 0 \;\; or \;\; (x – 1) = 0$

$\Rightarrow x = 0 \;\; or \;\; x = 1$

Since $x^{3} + x^{2} – ax + b$ is divisible by $x^{2} – x$

Hence, the zeros of $x^{2} – x$ will satisfy $x^{3} + x^{2} – ax + b$

$Therefore, (0)^{3} + 0 ^{2} – a(0) + b = 0$

$\Rightarrow b = 0$

And

$(1)^{3} + 1 ^{2} – a(1) + 0 = 0 \;\;\;\; [Since, b = 0]$

$\Rightarrow a = 2$

Q.16:  If $\alpha \;\; and \;\; \beta$ are the zeros of the polynomial $2x^{2} + 7x + 5$, write the value of $\alpha + \beta + \alpha \beta$ .

Sol:

By using the relationship between the zeros of the quadratic polynomial

We have,

Sum of zeros = $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$ and

Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$Therefore, \alpha + \beta = \frac{-7}{2} \;\; and \;\; \alpha \beta = \frac{-7}{5} + \frac{5}{2}$

Now, $Therefore, \alpha + \beta + \alpha \beta = \frac{-7}{2} + \frac{5}{2} = -1$

Q.17: State division algorithm for the polynomial.

Sol:

If f(x) and g(x) are two polynomial such that degree of f(x) is greater than degree of g(x) where g(x) $\neq$ 0, then there exist unique polynomial q(x) and r(x) such that

$f(x) = g(x) \times q(x) + r(x)$

Where r(x) = 0 or degree of r(x) $<$ degree of g(x).

Q.18: The sum of the zeros and the product of zeros of a Quadratic polynomial are $\frac{-1}{2} \;\; and \;\; -3$ respectively. Write the polynomial.

Sol:

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula

$x^{2}$ – (Sum of zeros ) x + Product of zeros

$\Rightarrow x^{2} – \left ( =\frac{1}{2} \right ) x + (-3)$

$\Rightarrow x^{2} + \frac{1}{2}x – 3$

Hence the required polynomial is $\Rightarrow x^{2} + \frac{1}{2}x – 3$

Q.19: Write the zeros of the Quadratic polynomial $f(x) = 6x^{2} -3$

Sol:

To find the zeros of the quadratic polynomial we will equate f(x) = 0

$Therefore, f(x) = 0$

$\Rightarrow 6x^{2} – 3 = 0$

$\Rightarrow 3 (2x^{2} – 1) = 0$

$\Rightarrow (2x^{2} – 1) = 0$

$\Rightarrow 2x^{2} = 1$

$\Rightarrow x^{2} = \frac{1}{2}$

$\Rightarrow x = \pm \frac{1}{\sqrt{2}}$

Hence, the zero of the quadratic polynomial $f(x) = 6x^{2} -3$ are $\frac{1}{\sqrt{2}} , \; – \frac{1}{\sqrt{2}}$

Q.20: Write the zeros of the Quadratic polynomial $f(x) = 4\sqrt{3} x^{2} + 5x – 2\sqrt{3}$

Sol:

To find the zeros of the quadratic polynomial we will equate f(x) = 0

$Therefore, f(x) = 0$

$\Rightarrow 4\sqrt{3} x^{2} + 5x – 2\sqrt{3} = 0$

$\Rightarrow 4\sqrt{3} x^{2} + 8x – 3x – 2\sqrt{3} = 0$

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

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

$\Rightarrow (4x – \sqrt{3}) = 0 \;\; or \;\; (\sqrt{3} x + 2) = 0$

$\Rightarrow x = \frac{-2}{\sqrt{3}} \;\; or \;\; x = \frac{\sqrt{3}}{4}$

Hence, the zeros of the quadratic polynomial f(x) = $f(x) = 4\sqrt{3} x^{2} + 5x – 2\sqrt{3}$ are $\frac{-2}{\sqrt{3}} \;\; or \;\; x = \frac{\sqrt{3}}{4}$

Q.21: Write the zeros of the Quadratic polynomial $f(x) = x^{2} – 5x + k$ such that $\alpha – \beta = 1$, find the value of k.

Sol:

By using the relationship between the zeros of the quadratic equation polynomial:

We have,

Sum of zeros =  $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$ and

Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$Therefore, \alpha + \beta = \frac{-(-5)}{1} \;\; and \;\; \alpha \beta = \frac{k}{1}$

$\Rightarrow \alpha + \beta = 5 \;\; and \;\; \alpha \beta = \frac{k}{1}$

Solving $\alpha – \beta = 1 \;\; and \;\; \alpha + \beta = 5$ we get

$\alpha = 3 \;\; and \;\; \beta = 2$

Substituting the values of $\alpha \beta = \frac{k}{1}$, we get

k = 6

Q.22: If $\alpha \;\; and \;\; \beta$ are zeros of the polynomial $f(x) = 6x^{2} + x – 2$, then find the value of $\frac{\alpha}{\beta } + \frac{\beta}{\alpha }$.

Sol;

By using the relationship between the zeros of the quadratic polynomials.

We have,

Sum of zeros =  $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$ and

Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$Therefore, \alpha + \beta = \frac{-(-1)}{6} \;\; and \;\; \alpha \beta = – \frac{1}{3}$

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha ^{2} + \beta^{2}}{\alpha \beta}$

$= \frac{ \alpha ^{2} + \beta^{2} + 2 \alpha \beta – 2 \alpha \beta }{\alpha \beta}$

$= \frac{ (\alpha ^{2} + \beta^{2})^{2} – 2 \alpha \beta }{\alpha \beta}$

$= \frac{ (- \frac{1}{6})^{2} – 2 \left ( -\frac{1}{6} \right )}{-\frac{1}{3}}$

$= \frac{ \frac{1}{36} + \frac{2}{3} } { \frac{2}{3}}$

$= – \frac{25} {12}$

Q.23: If $\alpha \;\; and \;\; \beta$ are zeros of the polynomial $f(x) = 5x^{2} – 7x + 1$, then find the value of $\frac{1}{\alpha } + \frac{1}{\beta }$.

Sol:

By using the relationship between the zeros of the quadratic polynomial.

We have,

Sum of zeros = $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$ and

Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$Therefore, \alpha + \beta = \frac{-(-7)}{5} \;\; and \;\; \alpha \beta = – \frac{1}{5}$

$\Rightarrow \alpha + \beta = \frac{7}{5} \;\; and \;\; \alpha \beta = \frac{1}{5}$

Now, $\frac{1}{\alpha } + \frac{1}{\beta } = \frac{\alpha +\beta }{\alpha \beta }$

$= \frac{\frac{7}{5}}{\frac{1}{5}}$ = 7

Q.24: If $\alpha \;\; and \;\; \beta$ are zeros of the polynomial $f(x) = x^{2} + x – 2$, then find the value of $\frac{1}{\alpha } – \frac{1}{\beta }$.

Sol:

By using the relationship between the zeros of the quadratic polynomial

We have,

Sum of zeros =  $\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}$ and

Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$\Rightarrow \alpha + \beta = \frac{-1}{1} \;\; and \;\; \alpha \beta = \frac{-2}{1}$

$\Rightarrow \alpha + \beta = -1 \;\; and \;\; \alpha \beta = -2$

Now, $\left ( \frac{1}{\alpha } – \frac{1}{\beta } \right )^{2} = \left ( \frac{\beta – \alpha }{\alpha \beta } \right )^{2}$

$= \frac{\left ( \alpha + \beta \right )^{2} – 4 \alpha \beta }{\left ( \alpha \beta \right )^{2}} \;\;\;\;\; \left [Since, \left ( \beta -\alpha \right )^{2} = \left ( \alpha + \beta \right )^{2} – 4\alpha \beta \right ]$

$= \frac{ ( -1 )^{2} – 4 (-2) }{(-2)^{2}} \;\;\;\;\; \left [ Since, \alpha + \beta = -1 \;\; and \;\; \alpha \beta = -2 \right ]$

$= \frac{1 + 8}{4}$

$= \frac{9}{4}$

$Since, \left ( \frac{1}{\alpha } – \frac{1}{\beta } \right )^{2} = \frac{9}{4}$

$\Rightarrow \frac{1}{\alpha } – \frac{1}{\beta } = \pm \frac{3}{2}$

Q.25: If the zeros of the polynomial $f(x) = x^{3} – 3x^{2} + x + 1$ are (a – b) , a and (a + b), find the value of a.

Sol:

By using the relationship between the zeros of the cubic polynomial:

We have,

Sum of zeros = $\frac{- (Coefficient \; of \; x )^{2}}{Coefficient \; of \; x^{3}}$

Therefore, a – b + a + a + b = $\frac{-(-3)}{1}$

$\Rightarrow 3a = 3$

$\Rightarrow a = 1$

Now, Product of zeros = $\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}$

$Therefore, (a – b) (a) (a + b) = \frac{-1}{1}$

$\Rightarrow (1 – b) (1) (1 + b) = -1 \left [ Since, a = 1 \right ]$

$\Rightarrow 1 – b^{2} = -1$

$\Rightarrow b^{2} = 2$

$\Rightarrow b = \pm \sqrt{2}$

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