# Question #4b3ff

Oct 11, 2017

$x = \frac{9}{3}$

#### Explanation:

Given -

$9 {x}^{2} - 25 = 0$

$9 {x}^{2} = 25$

${x}^{2} = \frac{25}{9}$
$x = \sqrt{\frac{25}{9}}$
$x = \frac{\sqrt{25}}{\sqrt{9}} = \frac{5}{3}$
$x = \frac{9}{3}$

Oct 11, 2017

$x = \setminus \frac{5}{3} , \setminus q \quad \setminus q \quad x = - \setminus \frac{5}{3}$

#### Explanation:

This is the standard form of a quadratic equation, which we can find the roots of by using the quadratic formula.

Rewriting the equation:

$9 {x}^{2} + 0 x - 25 = 0$

Our values to plug into the formula are:

• $a = 9$

• $b = 0$

• $c = - 25$

The formula is:

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plugging in for the variables:

$x = \setminus \frac{0 \setminus \pm \setminus \sqrt{{0}^{2} - 4 \left(9\right) \left(- 25\right)}}{2 \left(9\right)}$

$x = \setminus \frac{0 \setminus \pm \setminus \sqrt{0 + 900}}{18}$

$x = \setminus \frac{0 \setminus \pm \setminus \sqrt{900}}{18}$

$x = \setminus \frac{0 \setminus \pm 30}{18}$

$x = \setminus \frac{15}{18} , \setminus q \quad \setminus q \quad x = - \setminus \frac{30}{18}$

$\setminus \therefore x = \setminus \frac{5}{3} , \setminus q \quad \setminus q \quad x = - \setminus \frac{5}{3}$