Question

if alpha and beta are the zeros of polynomial such that alpha + beta is equal to 24 and Alpha minus beta is equal to 8 find the quadratic polynomial having alpha and beta as its zeros verify the relation​

Answer 1

AnswEr :

$\bf{\Large{\underline{\sf{Given\::}}}}$

If α and β are the zeros of polynomial such that α + β = 24 & α - β = 8.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The quadratic polynomial having α and β as its zeros verify the relation.

$\bf{\Large{\underline{\underline{\tt{\purple{Explanation\::}}}}}}$

We have;

• α + β = 24.......................(1)
• α - β = 8..........................(2)

Subtracting equation (1) and equation (2), we get;

$\implies\sf{\alpha \cancel{+\beta} +\alpha \cancel{-\beta} =24-8}\\\\\\\implies\sf{2\alpha \:=\:16}\\\\\\\implies\sf{\alpha \:=\:\cancel{\dfrac{16}{2}} }\\\\\\\implies\sf{\purple{\alpha \:=\:8}}\\\\\\\bf{\tt{Now,}}\\\\\sf{Putting\:the\:value\:of \alpha \:in\:equation(1)\:we\:get;}\\\\\implies\sf{8+\beta \:=\:24}\\\\\\\implies\sf{\beta \:=\:24\:-\:8}\\\\\\\implies\sf{\purple{\beta \:=\:16}}$

∴ The value is α = 8 & β = 16.

________________________________________

We know that sum of the zeroes :

$\implies\sf{\alpha +\beta \:=\:\dfrac{coefficient\:of\:x }{coefficient\:of\:x^{2}} }$

$\implies\sf{\purple{\alpha +\beta \:=\:24}}$

We know that product of zeroes :

$\implies\sf{\alpha \beta \:=\:\dfrac{constant\:term}{coefficient\:of\:x^{2} } }\\\\\\\implies\sf{\alpha \beta \:=\:8*16}\\\\\\\implies\sf{\purple{\alpha \beta \:=\:128}}$

Thus,

$\bigstar\:{\bf{\Large{\underline{\sf{The\:required\:quadratic\:equation\:is;}}}}}}$

$\mapsto\sf{x^{2} -(\alpha +\beta )x+\alpha \beta =0}\\\\\\\mapsto\sf{\purple{x^{2} -24x+128\:=\:0}}\:\:\:\:\:\:\:\:\:\:\:\:Ans.$

Answer 2

Answer:

AnswEr :

\bf{\Large{\underline{\sf{Given\::}}}}

Given:

If α and β are the zeros of polynomial such that α + β = 24 & α - β = 8.

\bf{\Large{\underline{\sf{To\:find\::}}}}

Tofind:

The quadratic polynomial having α and β as its zeros verify the relation.

\bf{\Large{\underline{\underline{\tt{\purple{Explanation\::}}}}}}

Explanation:

We have;

α + β = 24.......................(1)

α - β = 8..........................(2)

Subtracting equation (1) and equation (2), we get;

\begin{lgathered}\implies\sf{\alpha \cancel{+\beta} +\alpha \cancel{-\beta} =24-8}\\\\\\\implies\sf{2\alpha \:=\:16}\\\\\\\implies\sf{\alpha \:=\:\cancel{\dfrac{16}{2}} }\\\\\\\implies\sf{\purple{\alpha \:=\:8}}\\\\\\\bf{\tt{Now,}}\\\\\sf{Putting\:the\:value\:of \alpha \:in\:equation(1)\:we\:get;}\\\\\implies\sf{8+\beta \:=\:24}\\\\\\\implies\sf{\beta \:=\:24\:-\:8}\\\\\\\implies\sf{\purple{\beta \:=\:16}}\end{lgathered}

⟹α

+β

+α

−β

=24−8

⟹2α=16

⟹α=

2

16

⟹α=8

Now,

Puttingthevalueofαinequation(1)weget;

⟹8+β=24

⟹β=24−8

⟹β=16

∴ The value is α = 8 & β = 16.

________________________________________

We know that sum of the zeroes :

\implies\sf{\alpha +\beta \:=\:\dfrac{coefficient\:of\:x }{coefficient\:of\:x^{2}} }⟹α+β=

coefficientofx

2

coefficientofx

\implies\sf{\purple{\alpha +\beta \:=\:24}}⟹α+β=24

We know that product of zeroes :

\begin{lgathered}\implies\sf{\alpha \beta \:=\:\dfrac{constant\:term}{coefficient\:of\:x^{2} } }\\\\\\\implies\sf{\alpha \beta \:=\:8*16}\\\\\\\implies\sf{\purple{\alpha \beta \:=\:128}}\end{lgathered}

⟹αβ=

coefficientofx

2

constantterm

⟹αβ=8∗16

⟹αβ=128

Thus,

\begin{lgathered}\mapsto\sf{x^{2} -(\alpha +\beta )x+\alpha \beta =0}\\\\\\\mapsto\sf{\purple{x^{2} -24x+128\:=\:0}}\:\:\:\:\:\:\:\:\:\:\:\:Ans.\end{lgathered}

↦x

2

−(α+β)x+αβ=0

↦x

2

−24x+128=0Ans.