Solve ormula : c) 2x ^ 2 - 32 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { 2 x ^ { 2 } - 32 = 0 } \end{array} \right.$$

Solve for x

$x=4$
$x=-4$

Steps Using Difference of Squares

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps Using Difference of Squares

Divide both sides by $2$.

$$x^{2}-16=0$$

Consider $x^{2}-16$. Rewrite $x^{2}-16$ as $x^{2}-4^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve $x-4=0$ and $x+4=0$.

$$x=4$$ $$x=-4$$

Steps by Finding Square Root

Add $32$ to both sides. Anything plus zero gives itself.

$$2x^{2}=32$$

Divide both sides by $2$.

$$x^{2}=\frac{32}{2}$$

Divide $32$ by $2$ to get $16$.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4$$ $$x=-4$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an $x^{2}$ term but no $x$ term, can still be solved using the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$, once they are put in standard form: $ax^{2}+bx+c=0$.

$$2x^{2}-32=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $0$ for $b$, and $-32$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\times 2\left(-32\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{0±\sqrt{-8\left(-32\right)}}{2\times 2}$$

Multiply $-8$ times $-32$.

$$x=\frac{0±\sqrt{256}}{2\times 2}$$

Take the square root of $256$.

$$x=\frac{0±16}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{0±16}{4}$$

Now solve the equation $x=\frac{0±16}{4}$ when $±$ is plus. Divide $16$ by $4$.

$$x=4$$

Now solve the equation $x=\frac{0±16}{4}$ when $±$ is minus. Divide $-16$ by $4$.

$$x=-4$$

The equation is now solved.

$$x=4$$ $$x=-4$$