Question

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

Answer 1

Answer:

a = 1/x

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Answer 2

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