Let alpha, beta be two roots of the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$.We can rewrite this as $x^{2}+\sqrt{5} = -(20)^{\frac{1}{4}} x$.Squaring both

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) Given the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$.We can rewrite this as $x^{2}+\sqrt{5} = -(20)^{\frac{1}{4}} x$.Squaring both sides,we get $(x^{2}+\sqrt{5})^{2} = ((20)^{\frac{1}{4}} x)^{2}$.$x^{4} + 2\sqrt{5}x^{2} + 5 = \sqrt{20}x^{2}$.Since $\sqrt{20} = 2\sqrt{5}$,the equation becomes $x^{4} + 2\sqrt{5}x^{2} + 5 = 2\sqrt{5}x^{2}$.This simplifies to $x^{4} + 5 = 0$,or $x^{4} = -5$.Squaring again,$x^{8} = (-5)^{2} = 25$.Since $\alpha$ and $\beta$ are roots of the original equation,they satisfy $x^{4} = -5$,and thus $\alpha^{8} = 25$ and $\beta^{8} = 25$.Therefore,$\alpha^{8} + \beta^{8} = 25 + 25 = 50$.

Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to:

Similar Questions

The quadratic equation whose roots are the numbers having arithmetic mean $34$ and geometric mean $16$ is

If $(1 - p)$ is a root of the quadratic equation $x^2 + px + (1 - p) = 0$,then its roots are:

If the roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then:

If $x$ is real,then the minimum value of ${x^2} - 8x + 17$ is

If the equation $x^4+7x^3+18x^2+20x+8=0$ has a repeated root,then that repeated root is

Vedclass Test Series

Exam Paper Generator

Online Exam Module