Question #9042c

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Question #9042c

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Answer 1 · 2017-04-14T04:53:39

Solve the equation for y = 3.

Explanation:

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

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Question #9042c

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