Math, asked by deepa214, 1 year ago

if (a-b) and (a+b) are the roots 9f the polynomial f(x)=2x^3-6x^2+5x-7 then find a

Answers

Answered by bhoutik

sum of roots

we know that sum of roots of cubic polynomial is ( - coeff. of x^2) / coeff. of x^3)

let polynomial be f(x)

=) f(x) = 2x^3 -6x^2 +5x - 7

sum of roots = -(-6/2 )= 3

=) (a-b)+ a + (a+b) = 3

=) 3a = 3

=) a = 1

so value of a = 1

bhoutik: Where is s there is no such s in my answer.

bhoutik: I think you might be asking about the 3 ,am I right?

deepa214: ha s sry bimstkly i typed 3

bhoutik: Thats ( - coeff. of x^2) / coeff. of x^3)= -(-6/2 )= 3

bhoutik: Look carefully at the solution that you will definitely get it.

bhoutik: Than*

deepa214: on the side of adding the zeroes how xan we get 3a?

deepa214: can*

bhoutik: If you open (a-b)+a+(a+b) you will get a-b+a+a+b that is equal to 3a cuz b gets cancelled and add a+a+a =3a

bhoutik: This is also me here.

Answered by wardahd1234

Given:

$f(x) = 2 {x}^{3} - 6{x}^{2} + 5x - 7$

Now,

Sum of the zeros

$= - \frac{coefficient \: of \: {x}^{2} }{ \: coefficient \: of \: {x}^{3} }$

$(a - b) +a+ (a + b) \\ \\ = - \frac{( - 6)}{2} = 3$

$3a \: = 6$

$3a \: = 3 \\ \\ a = 1$

deepa214: how do we get 3a

bhoutik: If you open (a-b) + a +(a-b) you will get a-b+a+a+a+b

bhoutik: -b +b get cancelled and we get 3a

bhoutik: I hope it is cleared now.

bhoutik: This is also me whose answer you marked branliest i answered your query here also .

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