Find the values of x in terms of a in&nbsp;x2+(ax)2(x+a)2=3a2

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High School
Algebra
Algebra I
Quadratic function and equation

Jazmin Strong

Answered question

2022-04-07

Find the values of x in terms of a in $x^{2} + \frac{{(a x)}^{2}}{{(x + a)}^{2}} = 3 a^{2}$

Answer & Explanation

fallendreamsit2p

Beginner2022-04-08Added 14 answers

Step 1
$x^{2} + \frac{a^{2} x^{2}}{{(x + a)}^{2}} - \frac{2 a x^{2}}{a + x} + \frac{2 a x^{2}}{a + x} = 3 a^{2}$
${(x - \frac{a x}{a + x})}^{2} + \frac{2 a x^{2}}{a + x} = 3 a^{2}$
${(\frac{x^{2}}{a + x})}^{2} + \frac{2 a x^{2}}{a + x} - 3 a^{2} = 0$
${(\frac{x^{2}}{a + x})}^{2} + \frac{2 a x^{2}}{a + x} + a^{2} - a^{2} - 3 a^{2} = 0$
${(\frac{x^{2}}{a + x} + a)}^{2} - {(2 a)}^{2} = 0$
$(\frac{x^{2}}{a + x} - a) (\frac{x^{2}}{a + x} + 3 a) = 0$
and the rest is smooth:
The domain gives $x \neq - a$ , which gives $a \neq 0$ .
Step 2
Therefore, $x^{2} + 3 a x + 3 a^{2} = {(x + \frac{3 a}{2})}^{2} + \frac{3 a^{2}}{4} > 0,$
which says that the right polynomial does not give roots.
Also, $x^{2} - a x - a^{2} = 0$
gives ${\frac{a (1 + \sqrt{5})}{2}, \frac{a (1 - \sqrt{5})}{2}} .$
To solve an equation with parameter says to solve this equation for any values of the parameter.
Id est, we got the following answer.
If $a = 0$ so $o s l a s h$ ;
if $a \neq 0$ , so ${\frac{a (1 + \sqrt{5})}{2}, \frac{a (1 - \sqrt{5})}{2}} .$

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