Solve (a + 2) ^ 2 - 11a - 30 = 4(a + 25) | 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

$$(a+2)^{2}-11a-30=4(a+25)$$

Solve for a

$a=-7$
$a=18$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

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

$$a^{2}+4a+4-11a-30=4\left(a+25\right)$$

Combine $4a$ and $-11a$ to get $-7a$.

$$a^{2}-7a+4-30=4\left(a+25\right)$$

Subtract $30$ from $4$ to get $-26$.

$$a^{2}-7a-26=4\left(a+25\right)$$

Use the distributive property to multiply $4$ by $a+25$.

$$a^{2}-7a-26=4a+100$$

Subtract $4a$ from both sides.

$$a^{2}-7a-26-4a=100$$

Combine $-7a$ and $-4a$ to get $-11a$.

$$a^{2}-11a-26=100$$

Subtract $100$ from both sides.

$$a^{2}-11a-26-100=0$$

Subtract $100$ from $-26$ to get $-126$.

$$a^{2}-11a-126=0$$

To solve the equation, factor $a^{2}-11a-126$ using formula $a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-11$$ $$ab=-126$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-126$.

$$1,-126$$ $$2,-63$$ $$3,-42$$ $$6,-21$$ $$7,-18$$ $$9,-14$$

Calculate the sum for each pair.

$$1-126=-125$$ $$2-63=-61$$ $$3-42=-39$$ $$6-21=-15$$ $$7-18=-11$$ $$9-14=-5$$

The solution is the pair that gives sum $-11$.

$$a=-18$$ $$b=7$$

Rewrite factored expression $\left(a+a\right)\left(a+b\right)$ using the obtained values.

$$\left(a-18\right)\left(a+7\right)$$

To find equation solutions, solve $a-18=0$ and $a+7=0$.

$$a=18$$ $$a=-7$$

Steps Using Factoring By Grouping

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

$$a^{2}+4a+4-11a-30=4\left(a+25\right)$$

Combine $4a$ and $-11a$ to get $-7a$.

$$a^{2}-7a+4-30=4\left(a+25\right)$$

Subtract $30$ from $4$ to get $-26$.

$$a^{2}-7a-26=4\left(a+25\right)$$

Use the distributive property to multiply $4$ by $a+25$.

$$a^{2}-7a-26=4a+100$$

Subtract $4a$ from both sides.

$$a^{2}-7a-26-4a=100$$

Combine $-7a$ and $-4a$ to get $-11a$.

$$a^{2}-11a-26=100$$

Subtract $100$ from both sides.

$$a^{2}-11a-26-100=0$$

Subtract $100$ from $-26$ to get $-126$.

$$a^{2}-11a-126=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $a^{2}+aa+ba-126$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-11$$ $$ab=1\left(-126\right)=-126$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-126$.

$$1,-126$$ $$2,-63$$ $$3,-42$$ $$6,-21$$ $$7,-18$$ $$9,-14$$

Calculate the sum for each pair.

$$1-126=-125$$ $$2-63=-61$$ $$3-42=-39$$ $$6-21=-15$$ $$7-18=-11$$ $$9-14=-5$$

The solution is the pair that gives sum $-11$.

$$a=-18$$ $$b=7$$

Rewrite $a^{2}-11a-126$ as $\left(a^{2}-18a\right)+\left(7a-126\right)$.

$$\left(a^{2}-18a\right)+\left(7a-126\right)$$

Factor out $a$ in the first and $7$ in the second group.

$$a\left(a-18\right)+7\left(a-18\right)$$

Factor out common term $a-18$ by using distributive property.

$$\left(a-18\right)\left(a+7\right)$$

To find equation solutions, solve $a-18=0$ and $a+7=0$.

$$a=18$$ $$a=-7$$

Steps Using the Quadratic Formula

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

$$a^{2}+4a+4-11a-30=4\left(a+25\right)$$

Combine $4a$ and $-11a$ to get $-7a$.

$$a^{2}-7a+4-30=4\left(a+25\right)$$

Subtract $30$ from $4$ to get $-26$.

$$a^{2}-7a-26=4\left(a+25\right)$$

Use the distributive property to multiply $4$ by $a+25$.

$$a^{2}-7a-26=4a+100$$

Subtract $4a$ from both sides.

$$a^{2}-7a-26-4a=100$$

Combine $-7a$ and $-4a$ to get $-11a$.

$$a^{2}-11a-26=100$$

Subtract $100$ from both sides.

$$a^{2}-11a-26-100=0$$

Subtract $100$ from $-26$ to get $-126$.

$$a^{2}-11a-126=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-11$ for $b$, and $-126$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$a=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-126\right)}}{2}$$

Square $-11$.

$$a=\frac{-\left(-11\right)±\sqrt{121-4\left(-126\right)}}{2}$$

Multiply $-4$ times $-126$.

$$a=\frac{-\left(-11\right)±\sqrt{121+504}}{2}$$

Add $121$ to $504$.

$$a=\frac{-\left(-11\right)±\sqrt{625}}{2}$$

Take the square root of $625$.

$$a=\frac{-\left(-11\right)±25}{2}$$

The opposite of $-11$ is $11$.

$$a=\frac{11±25}{2}$$

Now solve the equation $a=\frac{11±25}{2}$ when $±$ is plus. Add $11$ to $25$.

$$a=\frac{36}{2}$$

Divide $36$ by $2$.

$$a=18$$

Now solve the equation $a=\frac{11±25}{2}$ when $±$ is minus. Subtract $25$ from $11$.

$$a=-\frac{14}{2}$$

Divide $-14$ by $2$.

$$a=-7$$

The equation is now solved.

$$a=18$$ $$a=-7$$

Steps for Completing the Square

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

$$a^{2}+4a+4-11a-30=4\left(a+25\right)$$

Combine $4a$ and $-11a$ to get $-7a$.

$$a^{2}-7a+4-30=4\left(a+25\right)$$

Subtract $30$ from $4$ to get $-26$.

$$a^{2}-7a-26=4\left(a+25\right)$$

Use the distributive property to multiply $4$ by $a+25$.

$$a^{2}-7a-26=4a+100$$

Subtract $4a$ from both sides.

$$a^{2}-7a-26-4a=100$$

Combine $-7a$ and $-4a$ to get $-11a$.

$$a^{2}-11a-26=100$$

Add $26$ to both sides.

$$a^{2}-11a=100+26$$

Add $100$ and $26$ to get $126$.

$$a^{2}-11a=126$$

Divide $-11$, the coefficient of the $x$ term, by $2$ to get $-\frac{11}{2}$. Then add the square of $-\frac{11}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-11a+\left(-\frac{11}{2}\right)^{2}=126+\left(-\frac{11}{2}\right)^{2}$$

Square $-\frac{11}{2}$ by squaring both the numerator and the denominator of the fraction.

$$a^{2}-11a+\frac{121}{4}=126+\frac{121}{4}$$

Add $126$ to $\frac{121}{4}$.

$$a^{2}-11a+\frac{121}{4}=\frac{625}{4}$$

Factor $a^{2}-11a+\frac{121}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(a-\frac{11}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{11}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

$$a-\frac{11}{2}=\frac{25}{2}$$ $$a-\frac{11}{2}=-\frac{25}{2}$$

Add $\frac{11}{2}$ to both sides of the equation.

$$a=18$$ $$a=-7$$