Solve (x-3)^2+5x^2=(2x+1)^2-x-1

$$( x - 3 ) ^ { 2 } + 5 x ^ { 2 } = ( 2 x + 1 ) ^ { 2 } - x - 1$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+5x^{2}=\left(2x+1\right)^{2}-x-1$$

Combine x^{2} and 5x^{2} to get 6x^{2}.

$$6x^{2}-6x+9=\left(2x+1\right)^{2}-x-1$$

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

$$6x^{2}-6x+9=4x^{2}+4x+1-x-1$$

Combine 4x and -x to get 3x.

$$6x^{2}-6x+9=4x^{2}+3x+1-1$$

Subtract 1 from 1 to get 0.

$$6x^{2}-6x+9=4x^{2}+3x$$

Subtract 4x^{2} from both sides.

$$6x^{2}-6x+9-4x^{2}=3x$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}-6x+9=3x$$

Subtract 3x from both sides.

$$2x^{2}-6x+9-3x=0$$

Combine -6x and -3x to get -9x.

$$2x^{2}-9x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\times 9=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite 2x^{2}-9x+9 as \left(2x^{2}-6x\right)+\left(-3x+9\right).

$$\left(2x^{2}-6x\right)+\left(-3x+9\right)$$

Factor out 2x in the first and -3 in the second group.

$$2x\left(x-3\right)-3\left(x-3\right)$$

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

$$\left(x-3\right)\left(2x-3\right)$$

To find equation solutions, solve x-3=0 and 2x-3=0.

$$x=3 x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+5x^{2}=\left(2x+1\right)^{2}-x-1$$

Combine x^{2} and 5x^{2} to get 6x^{2}.

$$6x^{2}-6x+9=\left(2x+1\right)^{2}-x-1$$

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

$$6x^{2}-6x+9=4x^{2}+4x+1-x-1$$

Combine 4x and -x to get 3x.

$$6x^{2}-6x+9=4x^{2}+3x+1-1$$

Subtract 1 from 1 to get 0.

$$6x^{2}-6x+9=4x^{2}+3x$$

Subtract 4x^{2} from both sides.

$$6x^{2}-6x+9-4x^{2}=3x$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}-6x+9=3x$$

Subtract 3x from both sides.

$$2x^{2}-6x+9-3x=0$$

Combine -6x and -3x to get -9x.

$$2x^{2}-9x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 9}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$x=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 2}$$

Add 81 to -72.

$$x=\frac{-\left(-9\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-\left(-9\right)±3}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±3}{4}$$

Now solve the equation x=\frac{9±3}{4} when ± is plus. Add 9 to 3.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{9±3}{4} when ± is minus. Subtract 3 from 9.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=3 x=\frac{3}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+5x^{2}=\left(2x+1\right)^{2}-x-1$$

Combine x^{2} and 5x^{2} to get 6x^{2}.

$$6x^{2}-6x+9=\left(2x+1\right)^{2}-x-1$$

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

$$6x^{2}-6x+9=4x^{2}+4x+1-x-1$$

Combine 4x and -x to get 3x.

$$6x^{2}-6x+9=4x^{2}+3x+1-1$$

Subtract 1 from 1 to get 0.

$$6x^{2}-6x+9=4x^{2}+3x$$

Subtract 4x^{2} from both sides.

$$6x^{2}-6x+9-4x^{2}=3x$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$2x^{2}-6x+9=3x$$

Subtract 3x from both sides.

$$2x^{2}-6x+9-3x=0$$

Combine -6x and -3x to get -9x.

$$2x^{2}-9x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-9x=-9$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=-\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=-\frac{9}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-\frac{9}{2}+\frac{81}{16}$$

Add -\frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{3}{4} x-\frac{9}{4}=-\frac{3}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=3 x=\frac{3}{2}$$