Solve 4x^2+4x+1=9x^2

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 9x^{2} from both sides.

$$4x^{2}+4x+1-9x^{2}=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}+4x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=4 ab=-5=-5$$

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. The only such pair is the system solution.

$$a=5 b=-1$$

Rewrite -5x^{2}+4x+1 as \left(-5x^{2}+5x\right)+\left(-x+1\right).

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

Factor out 5x in -5x^{2}+5x.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 9x^{2} from both sides.

$$4x^{2}+4x+1-9x^{2}=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}+4x+1=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-4±\sqrt{16+20}}{2\left(-5\right)}$$

Add 16 to 20.

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

Take the square root of 36.

$$x=\frac{-4±6}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-4±6}{-10}$$

Now solve the equation x=\frac{-4±6}{-10} when ± is plus. Add -4 to 6.

$$x=\frac{2}{-10}$$

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

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

Now solve the equation x=\frac{-4±6}{-10} when ± is minus. Subtract 6 from -4.

$$x=-\frac{10}{-10}$$

Divide -10 by -10.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 9x^{2} from both sides.

$$4x^{2}+4x+1-9x^{2}=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}+4x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}+4x=-1$$

Divide both sides by -5.

$$\frac{-5x^{2}+4x}{-5}=-\frac{1}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{4}{-5}x=-\frac{1}{-5}$$

Divide 4 by -5.

$$x^{2}-\frac{4}{5}x=-\frac{1}{-5}$$

Divide -1 by -5.

$$x^{2}-\frac{4}{5}x=\frac{1}{5}$$

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.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{5}\right)^{2}$$

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

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add \frac{1}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{3}{5} x-\frac{2}{5}=-\frac{3}{5}$$

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

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