How do you solve $9 x^{2} - 4 = 0$ $9x^2-4=0$ ?

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How do you solve $9 x^{2} - 4 = 0$ $9x^2-4=0$ ?

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Answer 1 · 2017-10-21T17:30:58

$x = \pm \frac{2}{3}$ $x=\pm\frac{2}{3}$

Explanation:

$9 x^{2} - 4 = 0$ $9x^2-4=0$

Move the $4$ $4$ to the RHS

$9 x^{2} = 4$ $9x^2=4$

Divide both sides by $9$ $9$

$x^{2} = \frac{4}{9}$ $x^2=\frac{4}{9}$

Take the square root of both sides

$\sqrt{x^{2}} = \pm \sqrt{\frac{4}{9}}$ $\sqrt{x^2}=\pm\sqrt{\frac{4}{9}}$

Simplify

$x = \pm \frac{2}{3}$ $x=\pm\frac{2}{3}$

$\thereforex=-\frac{2}{3},\qquad\qquad\qquad x=\frac{2}{3}$

Answer 2 · 2017-10-21T18:22:07

$x= +-sqrt(4/9$

In decimal $x = +-0.67$

Explanation:

$9x^2 - 4=0$
Let start by adding $color(red)(4)$ to both sides
$9x^2 cancel(- 4) cancelcolor(red)(+4)=0+color(red)(4)$
$9x^2 = 4$
Divide both sides by $color(green)(9)$
$(cancel9x^2)/cancelcolor(green)(9) = 4/color(green)(9)$
$x^2 = 4/9$
We need to take square root $sqrt$
$x= +-sqrt(4/9)$

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How do you solve $9 x^{2} - 4 = 0$ $9x^2-4=0$ ?

2 Answers

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How do you solve 9x^2-4=09x2−4=09x^2-4=0?

2 Answers

Explanation:

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How do you solve $9 x^{2} - 4 = 0$ $9x^2-4=0$ ?