Solve left(4x-5right)^2=left(5-3xright)^2+9/16x^2

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

Solve for x

$$x = \frac{160}{103} = 1\frac{57}{103} \approx 1.553398058$$

$$x=0$$

Steps Using Factoring

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

$$16x^{2}-40x+25=\left(5-3x\right)^{2}+\frac{9}{16}x^{2}$$

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

$$16x^{2}-40x+25=25-30x+9x^{2}+\frac{9}{16}x^{2}$$

Combine 9x^{2} and \frac{9}{16}x^{2} to get \frac{153}{16}x^{2}.

$$16x^{2}-40x+25=25-30x+\frac{153}{16}x^{2}$$

Subtract 25 from both sides.

$$16x^{2}-40x+25-25=-30x+\frac{153}{16}x^{2}$$

Subtract 25 from 25 to get 0.

$$16x^{2}-40x=-30x+\frac{153}{16}x^{2}$$

Add 30x to both sides.

$$16x^{2}-40x+30x=\frac{153}{16}x^{2}$$

Combine -40x and 30x to get -10x.

$$16x^{2}-10x=\frac{153}{16}x^{2}$$

Subtract \frac{153}{16}x^{2} from both sides.

$$16x^{2}-10x-\frac{153}{16}x^{2}=0$$

Combine 16x^{2} and -\frac{153}{16}x^{2} to get \frac{103}{16}x^{2}.

$$\frac{103}{16}x^{2}-10x=0$$

Factor out x.

$$x\left(\frac{103}{16}x-10\right)=0$$

To find equation solutions, solve x=0 and \frac{103x}{16}-10=0.

$$x=0 x=\frac{160}{103}$$

Steps Using the Quadratic Formula

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

$$16x^{2}-40x+25=\left(5-3x\right)^{2}+\frac{9}{16}x^{2}$$

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

$$16x^{2}-40x+25=25-30x+9x^{2}+\frac{9}{16}x^{2}$$

Combine 9x^{2} and \frac{9}{16}x^{2} to get \frac{153}{16}x^{2}.

$$16x^{2}-40x+25=25-30x+\frac{153}{16}x^{2}$$

Subtract 25 from both sides.

$$16x^{2}-40x+25-25=-30x+\frac{153}{16}x^{2}$$

Subtract 25 from 25 to get 0.

$$16x^{2}-40x=-30x+\frac{153}{16}x^{2}$$

Add 30x to both sides.

$$16x^{2}-40x+30x=\frac{153}{16}x^{2}$$

Combine -40x and 30x to get -10x.

$$16x^{2}-10x=\frac{153}{16}x^{2}$$

Subtract \frac{153}{16}x^{2} from both sides.

$$16x^{2}-10x-\frac{153}{16}x^{2}=0$$

Combine 16x^{2} and -\frac{153}{16}x^{2} to get \frac{103}{16}x^{2}.

$$\frac{103}{16}x^{2}-10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{103}{16} for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times \frac{103}{16}}$$

Take the square root of \left(-10\right)^{2}.

$$x=\frac{-\left(-10\right)±10}{2\times \frac{103}{16}}$$

The opposite of -10 is 10.

$$x=\frac{10±10}{2\times \frac{103}{16}}$$

Multiply 2 times \frac{103}{16}.

$$x=\frac{10±10}{\frac{103}{8}}$$

Now solve the equation x=\frac{10±10}{\frac{103}{8}} when ± is plus. Add 10 to 10.

$$x=\frac{20}{\frac{103}{8}}$$

Divide 20 by \frac{103}{8} by multiplying 20 by the reciprocal of \frac{103}{8}.

$$x=\frac{160}{103}$$

Now solve the equation x=\frac{10±10}{\frac{103}{8}} when ± is minus. Subtract 10 from 10.

$$x=\frac{0}{\frac{103}{8}}$$

Divide 0 by \frac{103}{8} by multiplying 0 by the reciprocal of \frac{103}{8}.

$$x=0$$

The equation is now solved.

$$x=\frac{160}{103} x=0$$

Steps for Completing the Square

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

$$16x^{2}-40x+25=\left(5-3x\right)^{2}+\frac{9}{16}x^{2}$$

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

$$16x^{2}-40x+25=25-30x+9x^{2}+\frac{9}{16}x^{2}$$

Combine 9x^{2} and \frac{9}{16}x^{2} to get \frac{153}{16}x^{2}.

$$16x^{2}-40x+25=25-30x+\frac{153}{16}x^{2}$$

Add 30x to both sides.

$$16x^{2}-40x+25+30x=25+\frac{153}{16}x^{2}$$

Combine -40x and 30x to get -10x.

$$16x^{2}-10x+25=25+\frac{153}{16}x^{2}$$

Subtract \frac{153}{16}x^{2} from both sides.

$$16x^{2}-10x+25-\frac{153}{16}x^{2}=25$$

Combine 16x^{2} and -\frac{153}{16}x^{2} to get \frac{103}{16}x^{2}.

$$\frac{103}{16}x^{2}-10x+25=25$$

Subtract 25 from both sides.

$$\frac{103}{16}x^{2}-10x=25-25$$

Subtract 25 from 25 to get 0.

$$\frac{103}{16}x^{2}-10x=0$$

Divide both sides of the equation by \frac{103}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{103}{16}x^{2}-10x}{\frac{103}{16}}=\frac{0}{\frac{103}{16}}$$

Dividing by \frac{103}{16} undoes the multiplication by \frac{103}{16}.

$$x^{2}+\left(-\frac{10}{\frac{103}{16}}\right)x=\frac{0}{\frac{103}{16}}$$

Divide -10 by \frac{103}{16} by multiplying -10 by the reciprocal of \frac{103}{16}.

$$x^{2}-\frac{160}{103}x=\frac{0}{\frac{103}{16}}$$

Divide 0 by \frac{103}{16} by multiplying 0 by the reciprocal of \frac{103}{16}.

$$x^{2}-\frac{160}{103}x=0$$

Divide -\frac{160}{103}, the coefficient of the x term, by 2 to get -\frac{80}{103}. Then add the square of -\frac{80}{103} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{160}{103}x+\left(-\frac{80}{103}\right)^{2}=\left(-\frac{80}{103}\right)^{2}$$

Square -\frac{80}{103} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{160}{103}x+\frac{6400}{10609}=\frac{6400}{10609}$$

Factor x^{2}-\frac{160}{103}x+\frac{6400}{10609}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{80}{103}\right)^{2}=\frac{6400}{10609}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{80}{103}\right)^{2}}=\sqrt{\frac{6400}{10609}}$$

Simplify.

$$x-\frac{80}{103}=\frac{80}{103} x-\frac{80}{103}=-\frac{80}{103}$$

Add \frac{80}{103} to both sides of the equation.

$$x=\frac{160}{103} x=0$$