Math, asked by pg5762177, 9 months ago

a(x^2+1)=(a^2+1)x,a≠0 by using formula ​

Answers

Answered by prashansa15
5

Answer:

a = 1/x

I hope it will help you

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Answered by Mankuthemonkey01
11

Answer

\sf x = a or \sf x = \frac{1}{a}

\rule{200}1

Explanation

Given

a(x² + 1) = (a² + 1)x, a ≠ 0

To find the value of x

a(x² + 1) = (a² + 1)x

→ ax² + a = (a² + 1)x

On Transposing, we get

→ ax² - (a² + 1)x + a = 0

Now, comparing it with the standard quadratic equation ax² + bx + c = 0, we get

a = a

b = - (a² + 1)

c = a

Now, by using quadratic formula we get

\sf x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Hence, the value of x would be

\sf x = \frac{ - (- (a^2 + 1)) \pm \sqrt{(a^2 + 1)^2 - 4(a)(a)}}{2a}

\sf x = \frac{a^2 + 1 \pm \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a}

\sf x = \frac{a^2 + 1 \pm \sqrt{a^4 + 1 - 2a^2}}{2a}

\sf x = \frac{a^2 + 1 \pm \sqrt{(a^2 - 1)^2}}{2a}

\sf x = \frac{a^2 + 1\pm (a^2 - 1)}{2a}

So,

\sf x = \frac{a^2 + 1 + a^2 - 1}{2a} or \sf x = \frac{a^2 + 1 - a^2 + 1}{2a}

\sf x = \frac{2a^2}{2a} or \sf x = \frac{2}{2a}

\sf x = a or \sf x = \frac{1}{a}

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