RS Aggarwal Class 10 Solutions Chapter 2 - Polynomials Ex 2B (2.2)

RS Aggarwal Class 10 Chapter 2 - Polynomials Ex 2B (2.2) Solutions Free PDF

The Class 10 Solutions of RS Aggarwal Maths for Chapter 2 is a complete set of study material that will help students to understand the basic concepts of Class 10 mathematics. While solving the exercise questions from the RS Aggarwal Maths textbook students can refer to these solutions if they get stuck in any of the questions. It is the best resource to prepare for mathematics examination. These solutions are sure to help the students to have a strong foundation and clear their final exams with excellent scores.

All the solutions are prepared by our highly skilled subject experts keeping in mind the difficulty level of the question. Practice these solutions once you are done covering the entire syllabus so that you can analyze your performance and get to know whether you have skipped any topic. It will also enhance your problem-solving skills and speed up your performance.

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

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

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

$\Rightarrow 1 – b^{2} = -1$

$\Rightarrow b^{2} = 2$

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

Key Features of RS Aggarwal Class 10 Solutions Chapter 2 – Polynomials  Ex 2B (2.2)

• The solutions are designed in a easy to understand way.
• All the exercise questions mentioned in the RS Aggarwal Class 10 Maths book are covered in these solutions.
• Practicing the solutions will help students to analyze their strength and weaknesses.
• One of the best revision material before the final exam.

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