Question #9042c

Apr 14, 2017

Solve the equation for y = 3.

Explanation:

Firstly, simplify ${x}^{- \frac{1}{2}} - 5 {x}^{- \frac{1}{4}} + 9 = 3$
Simplifying, we get $\frac{1}{\sqrt{x}} - \frac{5}{{x}^{\frac{1}{4}}} = 3 - 9 = - 6$
Simplify both fractions containing x,
$\frac{1}{\sqrt{x}} - \frac{5 \cdot \left({x}^{\frac{1}{4}}\right)}{{x}^{\frac{1}{4}} \cdot \left({x}^{\frac{1}{4}}\right)} = - 6$
$\frac{1}{\sqrt{x}} - \frac{5 {x}^{\frac{1}{4}}}{\sqrt{x}} = - 6$
$\frac{1 - 5 {x}^{\frac{1}{4}}}{\sqrt{x}} = - 6$
$\left(1 - 5 {x}^{\frac{1}{4}}\right) = - 6 \sqrt{x}$
$6 \sqrt{x} - 5 {x}^{\frac{1}{4}} + 1 = 0$
let ${a}^{2} = \sqrt{x} , a = {x}^{\frac{1}{4}}$
$6 {a}^{2} - 5 a + 1 = 0$
Factorise:
$6 {a}^{2} - 5 a + 1 = 0$
$\left(3 a - 1\right) \left(2 a - 1\right) = 0$
Thus $a = \frac{1}{3} \mathmr{and} a = \frac{1}{2}$
Since $a = {x}^{\frac{1}{4}} , {a}^{2} = \sqrt{x} = \frac{1}{9} , {a}^{4} = x , x = \frac{1}{81}$
and since ${a}^{4} = x , x = \frac{1}{2} ^ 4$
So $x = \frac{1}{16} \mathmr{and} x = \frac{1}{81}$
graph{x^(-1/2) - 5x^(-1/4) + 9 [-0.2597, 0.3647, 2.7107, 3.023]}