Solve x^2+1=2{left(5left(2x+1right)/5right)}^2

$${ x }^{ 2 } +1 = 2 { \left( \frac{ 5 \left( 2x+1 \right) }{ 5 } \right) }^{ 2 }$$

Solve for x

$$x=-\frac{1}{7}\approx -0.142857143$$

$$x=-1$$

Steps Using Factoring By Grouping

Cancel out 5 and 5.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{2}+1=2\left(4x^{2}+4x+1\right)$$

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$x^{2}+1=8x^{2}+8x+2$$

Subtract 8x^{2} from both sides.

$$x^{2}+1-8x^{2}=8x+2$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}+1=8x+2$$

Subtract 8x from both sides.

$$-7x^{2}+1-8x=2$$

Subtract 2 from both sides.

$$-7x^{2}+1-8x-2=0$$

Subtract 2 from 1 to get -1.

$$-7x^{2}-1-8x=0$$

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

$$-7x^{2}-8x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -7x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-7\left(-1\right)=7$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-7$$

Rewrite -7x^{2}-8x-1 as \left(-7x^{2}-x\right)+\left(-7x-1\right).

$$\left(-7x^{2}-x\right)+\left(-7x-1\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(7x+1\right)-\left(7x+1\right)$$

Factor out common term 7x+1 by using distributive property.

$$\left(7x+1\right)\left(-x-1\right)$$

To find equation solutions, solve 7x+1=0 and -x-1=0.

$$x=-\frac{1}{7} x=-1$$

Steps Using the Quadratic Formula

Cancel out 5 and 5.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{2}+1=2\left(4x^{2}+4x+1\right)$$

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$x^{2}+1=8x^{2}+8x+2$$

Subtract 8x^{2} from both sides.

$$x^{2}+1-8x^{2}=8x+2$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}+1=8x+2$$

Subtract 8x from both sides.

$$-7x^{2}+1-8x=2$$

Subtract 2 from both sides.

$$-7x^{2}+1-8x-2=0$$

Subtract 2 from 1 to get -1.

$$-7x^{2}-1-8x=0$$

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.

$$-7x^{2}-8x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, -8 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-7\right)\left(-1\right)}}{2\left(-7\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-7\right)\left(-1\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-\left(-8\right)±\sqrt{64+28\left(-1\right)}}{2\left(-7\right)}$$

Multiply 28 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64-28}}{2\left(-7\right)}$$

Add 64 to -28.

$$x=\frac{-\left(-8\right)±\sqrt{36}}{2\left(-7\right)}$$

Take the square root of 36.

$$x=\frac{-\left(-8\right)±6}{2\left(-7\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±6}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{8±6}{-14}$$

Now solve the equation x=\frac{8±6}{-14} when ± is plus. Add 8 to 6.

$$x=\frac{14}{-14}$$

Divide 14 by -14.

$$x=-1$$

Now solve the equation x=\frac{8±6}{-14} when ± is minus. Subtract 6 from 8.

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

Reduce the fraction \frac{2}{-14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{7}$$

The equation is now solved.

$$x=-1 x=-\frac{1}{7}$$

Steps for Completing the Square

Cancel out 5 and 5.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{2}+1=2\left(4x^{2}+4x+1\right)$$

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$x^{2}+1=8x^{2}+8x+2$$

Subtract 8x^{2} from both sides.

$$x^{2}+1-8x^{2}=8x+2$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}+1=8x+2$$

Subtract 8x from both sides.

$$-7x^{2}+1-8x=2$$

Subtract 1 from both sides.

$$-7x^{2}-8x=2-1$$

Subtract 1 from 2 to get 1.

$$-7x^{2}-8x=1$$

Divide both sides by -7.

$$\frac{-7x^{2}-8x}{-7}=\frac{1}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\left(-\frac{8}{-7}\right)x=\frac{1}{-7}$$

Divide -8 by -7.

$$x^{2}+\frac{8}{7}x=\frac{1}{-7}$$

Divide 1 by -7.

$$x^{2}+\frac{8}{7}x=-\frac{1}{7}$$

Divide \frac{8}{7}, the coefficient of the x term, by 2 to get \frac{4}{7}. Then add the square of \frac{4}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{7}x+\left(\frac{4}{7}\right)^{2}=-\frac{1}{7}+\left(\frac{4}{7}\right)^{2}$$

Square \frac{4}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=-\frac{1}{7}+\frac{16}{49}$$

Add -\frac{1}{7} to \frac{16}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=\frac{9}{49}$$

Factor x^{2}+\frac{8}{7}x+\frac{16}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{7}\right)^{2}=\frac{9}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{7}\right)^{2}}=\sqrt{\frac{9}{49}}$$

Simplify.

$$x+\frac{4}{7}=\frac{3}{7} x+\frac{4}{7}=-\frac{3}{7}$$

Subtract \frac{4}{7} from both sides of the equation.

$$x=-\frac{1}{7} x=-1$$