Solve 4\times-2y\times-2y=-4 | 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

$$4 \times -2y \times -2y=-4$$

Solve for y

$y=-\frac{1}{2}i=-0.5i$
$y=\frac{1}{2}i=0.5i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Multiply $y$ and $y$ to get $y^{2}$.

$$4\left(-2\right)y^{2}\left(-2\right)=-4$$

Multiply $4$ and $-2$ to get $-8$.

$$-8y^{2}\left(-2\right)=-4$$

Multiply $-8$ and $-2$ to get $16$.

$$16y^{2}=-4$$

Divide both sides by $16$.

$$y^{2}=\frac{-4}{16}$$

Reduce the fraction $\frac{-4}{16}$ to lowest terms by extracting and canceling out $4$.

$$y^{2}=-\frac{1}{4}$$

The equation is now solved.

$$y=\frac{1}{2}i$$ $$y=-\frac{1}{2}i$$

Steps Using the Quadratic Formula

Multiply $y$ and $y$ to get $y^{2}$.

$$4\left(-2\right)y^{2}\left(-2\right)=-4$$

Multiply $4$ and $-2$ to get $-8$.

$$-8y^{2}\left(-2\right)=-4$$

Multiply $-8$ and $-2$ to get $16$.

$$16y^{2}=-4$$

Add $4$ to both sides.

$$16y^{2}+4=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $16$ for $a$, $0$ for $b$, and $4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{0±\sqrt{0^{2}-4\times 16\times 4}}{2\times 16}$$

Square $0$.

$$y=\frac{0±\sqrt{-4\times 16\times 4}}{2\times 16}$$

Multiply $-4$ times $16$.

$$y=\frac{0±\sqrt{-64\times 4}}{2\times 16}$$

Multiply $-64$ times $4$.

$$y=\frac{0±\sqrt{-256}}{2\times 16}$$

Take the square root of $-256$.

$$y=\frac{0±16i}{2\times 16}$$

Multiply $2$ times $16$.

$$y=\frac{0±16i}{32}$$

Now solve the equation $y=\frac{0±16i}{32}$ when $±$ is plus.

$$y=\frac{1}{2}i$$

Now solve the equation $y=\frac{0±16i}{32}$ when $±$ is minus.

$$y=-\frac{1}{2}i$$

The equation is now solved.

$$y=\frac{1}{2}i$$ $$y=-\frac{1}{2}i$$