Question

if alpha and beta are the roots of the equation ax^2+bx+c=0 then the quadratic equation whose roots are alpha+beta and alpha×beta is
please tell the quadratic equation of this
to be marked as brainliest
​

Answer 1

ㅤ⋄Given :-

ㅤα, β are the roots of the Quadratic equation ax² + bx + c = 0

⋄ To find :-

The roots of the Quadratic equation whose root are

ㅤα+β ,αβ

ㅤㅤㅤㅤㅤ⋄ Solution:-

As we know that,

The Quadratic equation whose roots are ㅤα, β are

x² - (α+β) x +α β

So,

According to,

Sum of roots (α+ β) = -b/a

Product of roots (α β) = c/a

So, Now Required Quadratic Equation whose roots are , ㅤα+β ,αβ is :-

x² - (α+β+α β) x +(α+β)(αβ)=0

Substituting the values,

$\longrightarrow \: \sf x {}^{2} - \bigg(\dfrac{ - b}{a} + \dfrac{c}{a} \bigg) + \bigg( \dfrac{ - b}{a} \bigg) \bigg( \dfrac{c}{a} \bigg)=0$

$\longrightarrow \: \sf x {}^{2} - \bigg(\dfrac{ - b + c}{a} \bigg) + \bigg( \dfrac{ - b \times c}{a \times a} \bigg)=0$

$\longrightarrow \: \sf x {}^{2} - \bigg(\dfrac{ - b + c}{a} \bigg) + \bigg( \dfrac{ - bc}{a {}^{2} } \bigg)=0$

$\longrightarrow \: \sf x {}^{2} + \bigg(\dfrac{ b - c}{a} \bigg) - \bigg( \dfrac{ + bc}{a {}^{2} } \bigg) =0$

Taking L.C.M to the denominator .

$\longrightarrow \: \sf \: \dfrac{a {}^{2} {x {}^{} }^{2} +(b - c)a - bc }{a {}^{2} } = 0$

$\longrightarrow \: \sf \: {a {}^{2} {x {}^{} }^{2} +(b - c)a - bc } = 0 \times a {}^{2}$

$\longrightarrow \: \sf \: {a {}^{2} {x {}^{} }^{2} +(b - c)a - bc } = 0$

This is the required Quadratic equation whose roots are ㅤα+β ,αβ.