Solve (2x + 7/4) ^ 2 = (48x ^ 2 + 529)/16 | Equatix - Math Solver

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Question

$$(2x+\frac{7}{4})^{2}=\frac{48x^{2}+529}{16}$$

Solve for x

$x=-10$
$x=3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Multiply both sides of the equation by $16$.

$$16\left(2x+\frac{7}{4}\right)^{2}=48x^{2}+529$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+\frac{7}{4}\right)^{2}$.

$$16\left(4x^{2}+7x+\frac{49}{16}\right)=48x^{2}+529$$

Use the distributive property to multiply $16$ by $4x^{2}+7x+\frac{49}{16}$.

$$64x^{2}+112x+49=48x^{2}+529$$

Subtract $48x^{2}$ from both sides.

$$64x^{2}+112x+49-48x^{2}=529$$

Combine $64x^{2}$ and $-48x^{2}$ to get $16x^{2}$.

$$16x^{2}+112x+49=529$$

Subtract $529$ from both sides.

$$16x^{2}+112x+49-529=0$$

Subtract $529$ from $49$ to get $-480$.

$$16x^{2}+112x-480=0$$

Divide both sides by $16$.

$$x^{2}+7x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-30$. To find $a$ and $b$, set up a system to be solved.

$$a+b=7$$ $$ab=1\left(-30\right)=-30$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-30$.

$$-1,30$$ $$-2,15$$ $$-3,10$$ $$-5,6$$

Calculate the sum for each pair.

$$-1+30=29$$ $$-2+15=13$$ $$-3+10=7$$ $$-5+6=1$$

The solution is the pair that gives sum $7$.

$$a=-3$$ $$b=10$$

Rewrite $x^{2}+7x-30$ as $\left(x^{2}-3x\right)+\left(10x-30\right)$.

$$\left(x^{2}-3x\right)+\left(10x-30\right)$$

Factor out $x$ in the first and $10$ in the second group.

$$x\left(x-3\right)+10\left(x-3\right)$$

Factor out common term $x-3$ by using distributive property.

$$\left(x-3\right)\left(x+10\right)$$

To find equation solutions, solve $x-3=0$ and $x+10=0$.

$$x=3$$ $$x=-10$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by $16$.

$$16\left(2x+\frac{7}{4}\right)^{2}=48x^{2}+529$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+\frac{7}{4}\right)^{2}$.

$$16\left(4x^{2}+7x+\frac{49}{16}\right)=48x^{2}+529$$

Use the distributive property to multiply $16$ by $4x^{2}+7x+\frac{49}{16}$.

$$64x^{2}+112x+49=48x^{2}+529$$

Subtract $48x^{2}$ from both sides.

$$64x^{2}+112x+49-48x^{2}=529$$

Combine $64x^{2}$ and $-48x^{2}$ to get $16x^{2}$.

$$16x^{2}+112x+49=529$$

Subtract $529$ from both sides.

$$16x^{2}+112x+49-529=0$$

Subtract $529$ from $49$ to get $-480$.

$$16x^{2}+112x-480=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $16$ for $a$, $112$ for $b$, and $-480$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-112±\sqrt{112^{2}-4\times 16\left(-480\right)}}{2\times 16}$$

Square $112$.

$$x=\frac{-112±\sqrt{12544-4\times 16\left(-480\right)}}{2\times 16}$$

Multiply $-4$ times $16$.

$$x=\frac{-112±\sqrt{12544-64\left(-480\right)}}{2\times 16}$$

Multiply $-64$ times $-480$.

$$x=\frac{-112±\sqrt{12544+30720}}{2\times 16}$$

Add $12544$ to $30720$.

$$x=\frac{-112±\sqrt{43264}}{2\times 16}$$

Take the square root of $43264$.

$$x=\frac{-112±208}{2\times 16}$$

Multiply $2$ times $16$.

$$x=\frac{-112±208}{32}$$

Now solve the equation $x=\frac{-112±208}{32}$ when $±$ is plus. Add $-112$ to $208$.

$$x=\frac{96}{32}$$

Divide $96$ by $32$.

$$x=3$$

Now solve the equation $x=\frac{-112±208}{32}$ when $±$ is minus. Subtract $208$ from $-112$.

$$x=-\frac{320}{32}$$

Divide $-320$ by $32$.

$$x=-10$$

The equation is now solved.

$$x=3$$ $$x=-10$$

Steps for Completing the Square

Multiply both sides of the equation by $16$.

$$16\left(2x+\frac{7}{4}\right)^{2}=48x^{2}+529$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+\frac{7}{4}\right)^{2}$.

$$16\left(4x^{2}+7x+\frac{49}{16}\right)=48x^{2}+529$$

Use the distributive property to multiply $16$ by $4x^{2}+7x+\frac{49}{16}$.

$$64x^{2}+112x+49=48x^{2}+529$$

Subtract $48x^{2}$ from both sides.

$$64x^{2}+112x+49-48x^{2}=529$$

Combine $64x^{2}$ and $-48x^{2}$ to get $16x^{2}$.

$$16x^{2}+112x+49=529$$

Subtract $49$ from both sides.

$$16x^{2}+112x=529-49$$

Subtract $49$ from $529$ to get $480$.

$$16x^{2}+112x=480$$

Divide both sides by $16$.

$$\frac{16x^{2}+112x}{16}=\frac{480}{16}$$

Dividing by $16$ undoes the multiplication by $16$.

$$x^{2}+\frac{112}{16}x=\frac{480}{16}$$

Divide $112$ by $16$.

$$x^{2}+7x=\frac{480}{16}$$

Divide $480$ by $16$.

$$x^{2}+7x=30$$

Divide $7$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{2}$. Then add the square of $\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=30+\left(\frac{7}{2}\right)^{2}$$

Square $\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=30+\frac{49}{4}$$

Add $30$ to $\frac{49}{4}$.

$$x^{2}+7x+\frac{49}{4}=\frac{169}{4}$$

Factor $x^{2}+7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{13}{2}$$ $$x+\frac{7}{2}=-\frac{13}{2}$$

Subtract $\frac{7}{2}$ from both sides of the equation.

$$x=3$$ $$x=-10$$