Solve (4x+1)^2=6x^2+19

$$( 4 x + 1 ) ^ { 2 } = 6 x ^ { 2 } + 19$$

Solve for x

$$x = -\frac{9}{5} = -1\frac{4}{5} = -1.8$$

$$x=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=6x^{2}+19$$

Subtract 6x^{2} from both sides.

$$16x^{2}+8x+1-6x^{2}=19$$

Combine 16x^{2} and -6x^{2} to get 10x^{2}.

$$10x^{2}+8x+1=19$$

Subtract 19 from both sides.

$$10x^{2}+8x+1-19=0$$

Subtract 19 from 1 to get -18.

$$10x^{2}+8x-18=0$$

Divide both sides by 2.

$$5x^{2}+4x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=4 ab=5\left(-9\right)=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

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

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

Factor out 5x in the first and 9 in the second group.

$$5x\left(x-1\right)+9\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(5x+9\right)$$

To find equation solutions, solve x-1=0 and 5x+9=0.

$$x=1 x=-\frac{9}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=6x^{2}+19$$

Subtract 6x^{2} from both sides.

$$16x^{2}+8x+1-6x^{2}=19$$

Combine 16x^{2} and -6x^{2} to get 10x^{2}.

$$10x^{2}+8x+1=19$$

Subtract 19 from both sides.

$$10x^{2}+8x+1-19=0$$

Subtract 19 from 1 to get -18.

$$10x^{2}+8x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 8 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 10\left(-18\right)}}{2\times 10}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 10\left(-18\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-8±\sqrt{64-40\left(-18\right)}}{2\times 10}$$

Multiply -40 times -18.

$$x=\frac{-8±\sqrt{64+720}}{2\times 10}$$

Add 64 to 720.

$$x=\frac{-8±\sqrt{784}}{2\times 10}$$

Take the square root of 784.

$$x=\frac{-8±28}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-8±28}{20}$$

Now solve the equation x=\frac{-8±28}{20} when ± is plus. Add -8 to 28.

$$x=\frac{20}{20}$$

Divide 20 by 20.

$$x=1$$

Now solve the equation x=\frac{-8±28}{20} when ± is minus. Subtract 28 from -8.

$$x=-\frac{36}{20}$$

Reduce the fraction \frac{-36}{20} to lowest terms by extracting and canceling out 4.

$$x=-\frac{9}{5}$$

The equation is now solved.

$$x=1 x=-\frac{9}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=6x^{2}+19$$

Subtract 6x^{2} from both sides.

$$16x^{2}+8x+1-6x^{2}=19$$

Combine 16x^{2} and -6x^{2} to get 10x^{2}.

$$10x^{2}+8x+1=19$$

Subtract 1 from both sides.

$$10x^{2}+8x=19-1$$

Subtract 1 from 19 to get 18.

$$10x^{2}+8x=18$$

Divide both sides by 10.

$$\frac{10x^{2}+8x}{10}=\frac{18}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{8}{10}x=\frac{18}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{4}{5}x=\frac{18}{10}$$

Reduce the fraction \frac{18}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{4}{5}x=\frac{9}{5}$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\frac{9}{5}+\left(\frac{2}{5}\right)^{2}$$

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

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{9}{5}+\frac{4}{25}$$

Add \frac{9}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{49}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{7}{5} x+\frac{2}{5}=-\frac{7}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=1 x=-\frac{9}{5}$$