Question

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Answer 1

Step-by-step explanation:

Given bi-quadratic equation is:

$x^4+95x^2 -500=0$

Substituting $x^2=a,$ given bi-quadratic equation reduces in the form of following quadratic equation:

$a^2 +95a -500=0$

Let us factorize the above quadratic equation:

$a^2 +95a -500=0\\\therefore\: a^2 +100a -5a-500=0\\\therefore\: a(a +100) -5(a+100)=0\\\therefore\: (a +100)(a -5)=0\\\therefore\: (a +100)=0\:\: or \:\: (a -5)=0\\\therefore\: a = - 100\:\: or \:\: a = 5\\\\CASE\: (1)\:\\When\: a = - 100 \implies x^2 = - 100\\\therefore x=\pm \sqrt{-100}$

Since square root of a negative number cannot be found, so:

$x \neq \pm \sqrt{-100}$

$CASE\: (2)\:\\When\: a =5 \implies x^2 =5\\\therefore x=\pm \sqrt{5}$