Solve 7 = n ^ 2 + 1/4 * n ^ 2 | 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

$$7=n^{2}+\frac{1}{4}n^{2}$$

Solve for n

$n = \frac{2 \sqrt{35}}{5} \approx 2.366431913$
$n = -\frac{2 \sqrt{35}}{5} \approx -2.366431913$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Combine $n^{2}$ and $\frac{1}{4}n^{2}$ to get $\frac{5}{4}n^{2}$.

$$7=\frac{5}{4}n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{5}{4}n^{2}=7$$

Multiply both sides by $\frac{4}{5}$, the reciprocal of $\frac{5}{4}$.

$$n^{2}=7\times \frac{4}{5}$$

Multiply $7$ and $\frac{4}{5}$ to get $\frac{28}{5}$.

$$n^{2}=\frac{28}{5}$$

Take the square root of both sides of the equation.

$$n=\frac{2\sqrt{35}}{5}$$ $$n=-\frac{2\sqrt{35}}{5}$$

Steps Using the Quadratic Formula

Combine $n^{2}$ and $\frac{1}{4}n^{2}$ to get $\frac{5}{4}n^{2}$.

$$7=\frac{5}{4}n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{5}{4}n^{2}=7$$

Subtract $7$ from both sides.

$$\frac{5}{4}n^{2}-7=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $\frac{5}{4}$ for $a$, $0$ for $b$, and $-7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{0±\sqrt{0^{2}-4\times \frac{5}{4}\left(-7\right)}}{2\times \frac{5}{4}}$$

Square $0$.

$$n=\frac{0±\sqrt{-4\times \frac{5}{4}\left(-7\right)}}{2\times \frac{5}{4}}$$

Multiply $-4$ times $\frac{5}{4}$.

$$n=\frac{0±\sqrt{-5\left(-7\right)}}{2\times \frac{5}{4}}$$

Multiply $-5$ times $-7$.

$$n=\frac{0±\sqrt{35}}{2\times \frac{5}{4}}$$

Multiply $2$ times $\frac{5}{4}$.

$$n=\frac{0±\sqrt{35}}{\frac{5}{2}}$$

Now solve the equation $n=\frac{0±\sqrt{35}}{\frac{5}{2}}$ when $±$ is plus.

$$n=\frac{2\sqrt{35}}{5}$$

Now solve the equation $n=\frac{0±\sqrt{35}}{\frac{5}{2}}$ when $±$ is minus.

$$n=-\frac{2\sqrt{35}}{5}$$

The equation is now solved.

$$n=\frac{2\sqrt{35}}{5}$$ $$n=-\frac{2\sqrt{35}}{5}$$