The number of integral roots of the equation x4-x4-20-22 is

The correct option is C 2x^4+√(x^4 +20)=22 ⇒ nbsp;√(x^4 +20) = 22 x^4 Squaring on both sides, ⇒ x^4 +20 = (22 x^4)^2 Assuming x^4 = t ⇒ t+20 = (22 t)^2 ⇒ t ...

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Standard XIII

Mathematics

Higher Order Equations

The number of...

Question

The number of integral roots of the equation $x^{4} + \sqrt{x^{4} + 20} = 22$ is

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Solution

The correct option is B $2$
$x^{4} + \sqrt{x^{4} + 20} = 22 \Rightarrow \sqrt{x^{4} + 20} = 22 - x^{4}$
Squaring on both sides,
$\Rightarrow x^{4} + 20 = (22 - x^{4})^{2}$
Assuming $x^{4} = t$
$\Rightarrow t + 20 = (22 - t)^{2} \Rightarrow t^{2} - 45 t + 464 = 0 \Rightarrow (t - 29) (t - 16) = 0 \Rightarrow t = 16, 29 \Rightarrow x^{4} = 16, 29 \Rightarrow x = \pm 2, \pm (29)^{1 / 4}$

Hence, there are $2$ integral values of $x$ satisfying the equation.

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The number of integral roots of the equation x4+√x4+20=22 is

The correct option is B 2x4+√x4+20=22⇒√x4+20=22−x4 Squaring on both sides, ⇒x4+20=(22−x4)2 Assuming x4=t ⇒t+20=(22−t)2⇒t2−45t+464=0⇒(t−29)(t−16)=0⇒t=16,29⇒x4=16,29⇒x=±2,±(29)1/4 Hence, there are 2 integral values of x satisfying the equation.

The number of integral roots of the equation $x^{4} + \sqrt{x^{4} + 20} = 22$ is