Solve (5/4x)^2-(5-3/4x)^2=(2x-4)+9

$$( \frac { 5 } { 4 } x ) ^ { 2 } - ( 5 - \frac { 3 } { 4 } x ) ^ { 2 } = ( 2 x - 4 ) + 9$$

Solve for x

$$x = \frac{\sqrt{601} - 11}{4} \approx 3.378825336$$

$$x=\frac{-\sqrt{601}-11}{4}\approx -8.878825336$$

Steps Using the Quadratic Formula

Expand \left(\frac{5}{4}x\right)^{2}.

$$\left(\frac{5}{4}\right)^{2}x^{2}-\left(5-\frac{3}{4}x\right)^{2}=2x-4+9$$

Calculate \frac{5}{4} to the power of 2 and get \frac{25}{16}.

$$\frac{25}{16}x^{2}-\left(5-\frac{3}{4}x\right)^{2}=2x-4+9$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\frac{3}{4}x\right)^{2}.

$$\frac{25}{16}x^{2}-\left(25-\frac{15}{2}x+\frac{9}{16}x^{2}\right)=2x-4+9$$

To find the opposite of 25-\frac{15}{2}x+\frac{9}{16}x^{2}, find the opposite of each term.

$$\frac{25}{16}x^{2}-25+\frac{15}{2}x-\frac{9}{16}x^{2}=2x-4+9$$

Combine \frac{25}{16}x^{2} and -\frac{9}{16}x^{2} to get x^{2}.

$$x^{2}-25+\frac{15}{2}x=2x-4+9$$

Add -4 and 9 to get 5.

$$x^{2}-25+\frac{15}{2}x=2x+5$$

Subtract 2x from both sides.

$$x^{2}-25+\frac{15}{2}x-2x=5$$

Combine \frac{15}{2}x and -2x to get \frac{11}{2}x.

$$x^{2}-25+\frac{11}{2}x=5$$

Subtract 5 from both sides.

$$x^{2}-25+\frac{11}{2}x-5=0$$

Subtract 5 from -25 to get -30.

$$x^{2}-30+\frac{11}{2}x=0$$

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.

$$x^{2}+\frac{11}{2}x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{11}{2} for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{2}±\sqrt{\left(\frac{11}{2}\right)^{2}-4\left(-30\right)}}{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{11}{2}±\sqrt{\frac{121}{4}-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-\frac{11}{2}±\sqrt{\frac{121}{4}+120}}{2}$$

Add \frac{121}{4} to 120.

$$x=\frac{-\frac{11}{2}±\sqrt{\frac{601}{4}}}{2}$$

Take the square root of \frac{601}{4}.

$$x=\frac{-\frac{11}{2}±\frac{\sqrt{601}}{2}}{2}$$

Now solve the equation x=\frac{-\frac{11}{2}±\frac{\sqrt{601}}{2}}{2} when ± is plus. Add -\frac{11}{2} to \frac{\sqrt{601}}{2}.

$$x=\frac{\sqrt{601}-11}{2\times 2}$$

Divide \frac{-11+\sqrt{601}}{2} by 2.

$$x=\frac{\sqrt{601}-11}{4}$$

Now solve the equation x=\frac{-\frac{11}{2}±\frac{\sqrt{601}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{601}}{2} from -\frac{11}{2}.

$$x=\frac{-\sqrt{601}-11}{2\times 2}$$

Divide \frac{-11-\sqrt{601}}{2} by 2.

$$x=\frac{-\sqrt{601}-11}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{601}-11}{4} x=\frac{-\sqrt{601}-11}{4}$$

Steps for Completing the Square

Expand \left(\frac{5}{4}x\right)^{2}.

$$\left(\frac{5}{4}\right)^{2}x^{2}-\left(5-\frac{3}{4}x\right)^{2}=2x-4+9$$

Calculate \frac{5}{4} to the power of 2 and get \frac{25}{16}.

$$\frac{25}{16}x^{2}-\left(5-\frac{3}{4}x\right)^{2}=2x-4+9$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\frac{3}{4}x\right)^{2}.

$$\frac{25}{16}x^{2}-\left(25-\frac{15}{2}x+\frac{9}{16}x^{2}\right)=2x-4+9$$

To find the opposite of 25-\frac{15}{2}x+\frac{9}{16}x^{2}, find the opposite of each term.

$$\frac{25}{16}x^{2}-25+\frac{15}{2}x-\frac{9}{16}x^{2}=2x-4+9$$

Combine \frac{25}{16}x^{2} and -\frac{9}{16}x^{2} to get x^{2}.

$$x^{2}-25+\frac{15}{2}x=2x-4+9$$

Add -4 and 9 to get 5.

$$x^{2}-25+\frac{15}{2}x=2x+5$$

Subtract 2x from both sides.

$$x^{2}-25+\frac{15}{2}x-2x=5$$

Combine \frac{15}{2}x and -2x to get \frac{11}{2}x.

$$x^{2}-25+\frac{11}{2}x=5$$

Add 25 to both sides.

$$x^{2}+\frac{11}{2}x=5+25$$

Add 5 and 25 to get 30.

$$x^{2}+\frac{11}{2}x=30$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=30+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=30+\frac{121}{16}$$

Add 30 to \frac{121}{16}.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{601}{16}$$

Factor x^{2}+\frac{11}{2}x+\frac{121}{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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{601}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{601}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{\sqrt{601}}{4} x+\frac{11}{4}=-\frac{\sqrt{601}}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$x=\frac{\sqrt{601}-11}{4} x=\frac{-\sqrt{601}-11}{4}$$