How do you solve $2 x - 9 \sqrt{x} + 4 = 0$ $2x - 9\sqrt { x } + 4= 0$ ?

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How do you solve $2 x - 9 \sqrt{x} + 4 = 0$ $2x - 9\sqrt { x } + 4= 0$ ?

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Answer 1 · 2017-08-24T18:34:30

See below.

Explanation:

Move the term with the radical to the right hand side:

$2 x + 4 = 9 \sqrt{x}$ $2x + 4 = 9sqrt(x)$
square both sides and collect like terms:
$81 x = 4 x^{2} + 16 x + 16$ $81x = 4x^2 + 16x +16$
collect and equate to 0:
$4 x^{2} - 65 x + 16 = 0$ $4x^2 -65x + 16 = 0$
Factor:
$(4 x - 1) (x - 16) = 0$ $(4x - 1)(x -16) = 0$
solution: $x = 16, x = \frac{1}{4}$ $x = 16 , x = 1/4$

Answer 2 · 2017-08-24T18:43:38

$x = 16$ $x=16" "$ or $x = \frac{1}{4}$ $" "x = 1/4$

Explanation:

Given:

$2 x - 9 \sqrt{x} + 4 = 0$ $2x-9sqrt(x)+4 = 0$

Let $t = \sqrt{x}$ $t=sqrt(x)$ and find:

$0 = 8 (2 x - 9 \sqrt{x} + 4)$ $0 = 8(2x-9sqrt(x)+4)$

$0 = 8 (2 t^{2} - 9 t + 4)$ $color(white)(0) = 8(2t^2-9t+4)$

$0 = 16 t^{2} - 72 t + 32$ $color(white)(0) = 16t^2-72t+32$

$0 = {(4 t)}^{2} - 2 (4 t) (9) + 9^{2} - 49$ $color(white)(0) = (4t)^2-2(4t)(9)+9^2-49$

$0 = {(4 t - 9)}^{2} - 7^{2}$ $color(white)(0) = (4t-9)^2-7^2$

$0 = ((4 t - 9) - 7) ((4 t - 9) + 7)$ $color(white)(0) = ((4t-9)-7)((4t-9)+7)$

$0 = (4 t - 16) (4 t - 2)$ $color(white)(0) = (4t-16)(4t-2)$

$0 = 8 (t - 4) (2 t - 1)$ $color(white)(0) = 8(t-4)(2t-1)$

So:

$t = 4$ $t=4" "$ or $t = \frac{1}{2}$ $" "t=1/2$

Then:

$x = t^{2} = 4^{2} = 16$ $x = t^2 = 4^2 = 16" "$ or $x = t^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$ $" "x = t^2 = (1/2)^2 = 1/4$

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How do you solve $2 x - 9 \sqrt{x} + 4 = 0$ $2x - 9\sqrt { x } + 4= 0$ ?

2 Answers

Explanation:

Explanation:

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