Solve x^2+2x+1=49

$${ x }^{ 2 } +2x+1=49$$

Solve for x

$$x=-8$$

$$x=6$$

Steps Using Factoring

Subtract 49 from both sides.

$$x^{2}+2x+1-49=0$$

Subtract 49 from 1 to get -48.

$$x^{2}+2x-48=0$$

To solve the equation, factor x^{2}+2x-48 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 2.

$$a=-6 b=8$$

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

$$\left(x-6\right)\left(x+8\right)$$

To find equation solutions, solve x-6=0 and x+8=0.

$$x=6 x=-8$$

Steps Using Factoring By Grouping

Subtract 49 from both sides.

$$x^{2}+2x+1-49=0$$

Subtract 49 from 1 to get -48.

$$x^{2}+2x-48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-48\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 2.

$$a=-6 b=8$$

Rewrite x^{2}+2x-48 as \left(x^{2}-6x\right)+\left(8x-48\right).

$$\left(x^{2}-6x\right)+\left(8x-48\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-6\right)+8\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+8\right)$$

To find equation solutions, solve x-6=0 and x+8=0.

$$x=6 x=-8$$

Steps Using the Quadratic Formula

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.

$$x^{2}+2x+1=49$$

Subtract 49 from both sides of the equation.

$$x^{2}+2x+1-49=49-49$$

Subtracting 49 from itself leaves 0.

$$x^{2}+2x+1-49=0$$

Subtract 49 from 1.

$$x^{2}+2x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-48\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-48\right)}}{2}$$

Multiply -4 times -48.

$$x=\frac{-2±\sqrt{4+192}}{2}$$

Add 4 to 192.

$$x=\frac{-2±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-2±14}{2}$$

Now solve the equation x=\frac{-2±14}{2} when ± is plus. Add -2 to 14.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{-2±14}{2} when ± is minus. Subtract 14 from -2.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=6 x=-8$$

Steps for Completing the Square

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+1=7 x+1=-7$$

Subtract 1 from both sides of the equation.

$$x=6 x=-8$$