Solve the area of the e area of circle z = 16 | 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

$$\left. \begin{array} { l } { 6 2 + 4 y ^ { 2 } = 16 } \end{array} \right.$$

Solve for y

$y=1$
$y=-1$

Steps Using Difference of Squares

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps Using Difference of Squares

Multiply $6$ and $2$ to get $12$.

$$12+4y^{2}=16$$

Subtract $16$ from both sides.

$$12+4y^{2}-16=0$$

Subtract $16$ from $12$ to get $-4$.

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

Divide both sides by $4$.

$$-1+y^{2}=0$$

Consider $-1+y^{2}$. Rewrite $-1+y^{2}$ as $y^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.

$$\left(y-1\right)\left(y+1\right)=0$$

To find equation solutions, solve $y-1=0$ and $y+1=0$.

$$y=1$$ $$y=-1$$

Steps by Finding Square Root

Multiply $6$ and $2$ to get $12$.

$$12+4y^{2}=16$$

Subtract $12$ from both sides.

$$4y^{2}=16-12$$

Subtract $12$ from $16$ to get $4$.

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

Divide both sides by $4$.

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

Divide $4$ by $4$ to get $1$.

$$y^{2}=1$$

Take the square root of both sides of the equation.

$$y=1$$ $$y=-1$$

Steps Using the Quadratic Formula

Multiply $6$ and $2$ to get $12$.

$$12+4y^{2}=16$$

Subtract $16$ from both sides.

$$12+4y^{2}-16=0$$

Subtract $16$ from $12$ to get $-4$.

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

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: $ax^{2}+bx+c=0$.

$$4y^{2}-4=0$$

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

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

Square $0$.

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

Multiply $-4$ times $4$.

$$y=\frac{0±\sqrt{-16\left(-4\right)}}{2\times 4}$$

Multiply $-16$ times $-4$.

$$y=\frac{0±\sqrt{64}}{2\times 4}$$

Take the square root of $64$.

$$y=\frac{0±8}{2\times 4}$$

Multiply $2$ times $4$.

$$y=\frac{0±8}{8}$$

Now solve the equation $y=\frac{0±8}{8}$ when $±$ is plus. Divide $8$ by $8$.

$$y=1$$

Now solve the equation $y=\frac{0±8}{8}$ when $±$ is minus. Divide $-8$ by $8$.

$$y=-1$$

The equation is now solved.

$$y=1$$ $$y=-1$$