Question

Yeh question jaldi se solve kare yaar... Plss...

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Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\red{\implies\:144}$

$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• $\:f(x)\:=\:x^2-4x+3$

• $\alpha\:and\beta\:are\:zeros$

$\large{\underline{\underline{\mathfrak{\bf{Find\:here:-}}}}}$.

• $\:value\:of\:(\alpha^4\beta^2+\alpha^2\beta^4)$

$\large{\underline{\underline{\mathfrak{\bf{Explanation:-}}}}}$.

We know that's ,

$\large\boxed{\:Sum\:of\:Zeros\:=\frac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}$

$\implies\:(\alpha+\beta)\:=\frac{-(-4)}{1}$

$\implies\:(\alpha+\beta)\:=\: 4$....(1)

Now,

Squaring Both side ,

$\implies\:(\alpha+\beta)^2\:=\:(4)^2$

$\implies\:(\alpha+\beta)^2\:=\:16$.....(2)

Again,

$\large\boxed{\:(Product\:of\:Zeros)\:=\frac{(Constant\:part)}{(coefficient\:of\:x^2)}}$

$\implies\:(\alpha\beta)\:=\frac{(3)}{1}$

$\implies\:(\alpha\beta)\:=\;3$.....(3)

We Find Here,

$\implies\:(\alpha^4\beta^2+\alpha^2\beta^4)$

$\implies\:\alpha^2\beta^2(\alpha^2+\beta^2)$

Keep Value by equation (2) and (3),

$\implies\:(3)^2×(16)$

$\implies\:9×16$

$\implies\boxed{\:144}$

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