Solve (sqrt{19x+85})^2=(x+7)^2

$$( \sqrt { 19 x + 85 } ) ^ { 2 } = ( x + 7 ) ^ { 2 }$$

Solve for x

$$x=-4$$

$$x=9$$

Steps Using Factoring By Grouping

Calculate \sqrt{19x+85} to the power of 2 and get 19x+85.

$$19x+85=\left(x+7\right)^{2}$$

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

$$19x+85=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$19x+85-x^{2}=14x+49$$

Subtract 14x from both sides.

$$19x+85-x^{2}-14x=49$$

Combine 19x and -14x to get 5x.

$$5x+85-x^{2}=49$$

Subtract 49 from both sides.

$$5x+85-x^{2}-49=0$$

Subtract 49 from 85 to get 36.

$$5x+36-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+5x+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=5 ab=-36=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=9 b=-4$$

Rewrite -x^{2}+5x+36 as \left(-x^{2}+9x\right)+\left(-4x+36\right).

$$\left(-x^{2}+9x\right)+\left(-4x+36\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-9\right)-4\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-4\right)$$

To find equation solutions, solve x-9=0 and -x-4=0.

$$x=9 x=-4$$

Steps Using the Quadratic Formula

Calculate \sqrt{19x+85} to the power of 2 and get 19x+85.

$$19x+85=\left(x+7\right)^{2}$$

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

$$19x+85=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$19x+85-x^{2}=14x+49$$

Subtract 14x from both sides.

$$19x+85-x^{2}-14x=49$$

Combine 19x and -14x to get 5x.

$$5x+85-x^{2}=49$$

Subtract 49 from both sides.

$$5x+85-x^{2}-49=0$$

Subtract 49 from 85 to get 36.

$$5x+36-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+5x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 36}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 36}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 36}}{2\left(-1\right)}$$

Multiply 4 times 36.

$$x=\frac{-5±\sqrt{25+144}}{2\left(-1\right)}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±13}{-2}$$

Now solve the equation x=\frac{-5±13}{-2} when ± is plus. Add -5 to 13.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{-5±13}{-2} when ± is minus. Subtract 13 from -5.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-4 x=9$$

Steps for Completing the Square

Calculate \sqrt{19x+85} to the power of 2 and get 19x+85.

$$19x+85=\left(x+7\right)^{2}$$

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

$$19x+85=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$19x+85-x^{2}=14x+49$$

Subtract 14x from both sides.

$$19x+85-x^{2}-14x=49$$

Combine 19x and -14x to get 5x.

$$5x+85-x^{2}=49$$

Subtract 85 from both sides.

$$5x-x^{2}=49-85$$

Subtract 85 from 49 to get -36.

$$5x-x^{2}=-36$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+5x=-36$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=-\frac{36}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=-\frac{36}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=-\frac{36}{-1}$$

Divide -36 by -1.

$$x^{2}-5x=36$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=36+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=36+\frac{25}{4}$$

Add 36 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{169}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{13}{2} x-\frac{5}{2}=-\frac{13}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=9 x=-4$$