Question

.If the zeroes of the polynomial are 3x2 − 5x + 2 are a+ b and a- b, find a and b.

Answer 1

given ,the zeros of the polynomial are a+b and a-b

we know that sum of zeros =- x coefficient /x² coefficient

=2a=5/3

=a=5/6


product of roots = constant term /x² coefficient

=a²-b²=2/3

=(5/6)²-b²=2/3

=25/36-b²=2/3

=25/36-2/3=b²

=25-24/36=b²

=1/36=b²

b=±1/6



I hope this will help u ;)

Answer 2

Answer:

The roots of the polynomial are

 $a=\dfrac{5}{6} ~\mbox{and}~b=\dfrac{1}{6}$

Step-by-step explanation:

Given, the polynomial $3x^2 -5x + 2$

And the zeros of the polynomial are $a+b$ and $a-b$.

So, let $\alpha=a+b$ and $\beta=a-b$.  

For a polynomial of the form, $ax^2 +bx + c$,

• the sum of the roots of an equation is $-{b}/{a}$
• the product of the roots of an equation is ${c}/{a}$

The sum of the roots of the given polynomial, $3x^2 -5x + 2$ is

$\alpha+\beta=-\dfrac{(-5)}{3}=\dfrac{5}{3}$

$a+b+a-b=\dfrac{5}{3}\implies 2a =\dfrac{5}{3}\implies a =\dfrac{5}{3\times 2}=\dfrac{5}{6}$

The product of the roots of the given polynomial, $3x^2 -5x + 2$ is

$\alpha\times \beta =\dfrac{2}{3}\implies a+ b\times a- b =\dfrac{2}{3}\implies a^2 - b ^2=\dfrac{2}{3}$

Substituting the value of a,

$a^2 - b ^2=\dfrac{2}{3}\implies \bigg(\dfrac{5}{6}\bigg)^2 - b ^2=\dfrac{2}{3}$

$\implies b ^2=\dfrac{25-24}{36} = \dfrac{1}{36}$

$\implies b =\dfrac{1}{\sqrt{36} }=\pm \dfrac{1}{6 }$

Therefore, the roots of the polynomial are

$a=\dfrac{5}{6} ~\mbox{and}~b=\dfrac{1}{6}$

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