Ex.2.4 Q3 Polynomials Solution - NCERT Maths Class 10

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Question

If the zeroes of the polynomial \(x^{3}-3 x^{2}+x+1\) are \(a-b, \;a,\;a+b,\) find \(a \) and \(b \).

 Video Solution
Polynomials
Ex 2.4 | Question 3

Text Solution

What is known?

Zeroes of the polynomial \(x^{3}-3 x^{2}+x+1\) are \(a- b, \; a,\; a+b\)

What is unknown?

\(a \) and \(b \).

Reasoning:

First compare the given polynomial with the general equation of the cubic polynomial \(p{x^3} + q{x^2} + rx + t\) and you will get the values of \(p, \;q,\; r \) and \(t.\)

Now put this value in the equation of sum of zeroes and product of zeroes, you will get the value of \(a \) and \(b \).

Steps:

\[p\left( x \right) = {x^3} - 3{x^2} + x{{ }} + 1\]

Zeroes are \(a - b,{{ }}a,\;a + b.\)

On comparing the given polynomial with \(p{x^3} + q{x^2} + rx + t\)

we get,

\(p = 1, \;q = - 3,\;r = 1 \) and \(t = 1\)

Sum of zeroes

\[\begin{align}&= a - b + a + a + b\\& = \frac{{ - {\text{coeficient of }}{x^2}}}{{{{ }}{\text{coeficient of }}{x^3}}} = \frac{{ - q}}{p}\\{\frac{{ - q}}{p}}&={ 3{{a}}}\\{\frac{{ - ( - 3)}}{1}} &= {3{{a}}}\\a{{ }}& = {{ }}1\end{align}\]

Since the value of \(a\) is found to be \(1,\) the zeroes are \(1 - b,\;1,\;1 + b\)

Multiplication of zeroes

\[\begin{align}&= 1\left( {1 - b} \right)\left( {1 + b} \right)\\& = \frac{{ - {\text{constant term}}}}{\text{coeficient of }{x^3}}\\&= \frac{{ - t}}{p}\\{\frac{{ - t}}{p}} &= {1-{{ }}{b^2}}\\\frac{{ - 1}}{1} &= {1-{{ }}{b^2}}\\1 - {b^2}& = 1\\1 + 1 &= {b^2}\\{b^2} &= 2\end{align}\]

\(b = \pm \sqrt 2 .\) Hence, \(a = 1, \;b = \sqrt 2 \) or \(- \sqrt 2\)

  
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