Question

alpha and beta is are the zeros of the quadratic polynomial f(x) = x²-x-4, then find the value of alpha⁴+ beta⁴​

Answer 1

AnswEr :

49.

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

α and β is the zeroes of the quadratic polynomial f(x) = x² - x - 4.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The value of $\alpha ^{4} +\beta ^{4}$

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

We have quadratic polynomial:f(x) = x² - x - 4 compared with ax² + bx+c=0

• a = 1
• b = -1
• c = -4

$\bf{\underline{\underline{\bf{According\:to\:the\:question\::}}}}}$

$\mathcal{\underline{\orange{SUM\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{-(-1)}{1} }\\\\\\\mapsto\sf{\red{\alpha +\beta =1}}$

$\mathcal{\underline{\orange{PRODUCT\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{-4}{1} }\\\\\\\mapsto\sf{\red{\alpha \times \beta =-4}}$

Now;

$\mapsto\sf{\orange{\alpha ^{4} +\beta ^{4} =\big(\alpha ^{2} +\beta ^{2} \big)^{2} -2\alpha ^{2} \beta ^{2} }}\\\\\\\mapsto\sf{\alpha ^{4} +\beta ^{4}=\big[(\alpha +\beta )^{2} -2\alpha \beta \big]^{2} -2(\alpha \beta )^{2} }$

Putting the value of the zeroes in above formula :

$\mapsto\sf{\big[(-1)^{2} -2(-4)\big]^{2} -2(-4)^{2} }\\\\\\\mapsto\sf{\big[(1+8)^{2} -2\times 16\big]}\\\\\\\mapsto\sf{9^{2} -32}\\\\\\\mapsto\sf{81-32}\\\\\\\mapsto\sf{\red{49}}$

Thus;

$\underbrace{\sf{The\:value\:of\:\alpha ^{4}+\beta ^{4} \:=\:\pink{49}}}}}$

Answer 2

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❚ QuEstiOn ❚

$\alpha$ and $\beta$ are the zeros of the quadratic polynomial

➩ f(x) = X²-X-4,

Then find the value of $\alpha^4 +\beta^4$

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❚ ANsWeR ❚

✺ Given :

• Polynomial = f(x) = X²-X-4
• Zeroes are = $\alpha$ and $\beta$

✺ To FinD :

• $\alpha^4 +\beta^4$ = ?

✺ Explanation :

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We know that, Polynomial of having zeroes $\alpha$ and $\beta$ can be written as ,

✏ $\ \ {X^2-(\alpha+\beta)X+\alpha\beta=0}$

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Now ,

$\ \ {f(x)=X^2-X-4}$ Comparing this equation with $\ \ {X^2-(\alpha+\beta)X+\alpha\beta}$ we get ,

:$\longrightarrow (\alpha+\beta)=1$

:$\longrightarrow (\alpha\beta)=-4$

$\therefore\:\:\ \ {\alpha^4 +\beta^4}$

$\implies\ \ {(\alpha^2)^2 +(\beta^2)^2}$

$\implies\ \ {(\alpha^2 +\beta^2)^2-2\alpha^2\beta^2}$

$\implies\ \ {(\alpha^2 +\beta^2)^2-2(\alpha\beta)^2}$

$\implies\ \ {\{(\alpha +\beta)^2-2\alpha\beta\}^2-2(\alpha\beta)^2}$

Now , putting the values , $(\alpha+\beta)=1$ and $\alpha\beta=-4$

$\implies\ \ {\{(1)^2-2(-4)\}^2-2(-4)^2}$

$\implies\ \ {(1+8)^2-32}$

$\implies\ \ {9^2-32}$

$\implies\ \ {81-32}$

$\implies\ \ {49}$

$\therefore\:\:\boxed{\red{\ \ {(\alpha^4 +\beta^4)=49}}}$

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❏ Formula UseD :-

$\blacksquare \:\:\:\:\boxed{A^2+B^2=(A+B)^2-2AB}$

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