Solve (5x-4)^2+-(4x+1)^2=0

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

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$25x^{2}-40x+16-\left(4x+1\right)^{2}=0$$

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

$$25x^{2}-40x+16-\left(16x^{2}+8x+1\right)=0$$

To find the opposite of 16x^{2}+8x+1, find the opposite of each term.

$$25x^{2}-40x+16-16x^{2}-8x-1=0$$

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

$$9x^{2}-40x+16-8x-1=0$$

Combine -40x and -8x to get -48x.

$$9x^{2}-48x+16-1=0$$

Subtract 1 from 16 to get 15.

$$9x^{2}-48x+15=0$$

Divide both sides by 3.

$$3x^{2}-16x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-15 b=-1$$

Rewrite 3x^{2}-16x+5 as \left(3x^{2}-15x\right)+\left(-x+5\right).

$$\left(3x^{2}-15x\right)+\left(-x+5\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(3x-1\right)$$

To find equation solutions, solve x-5=0 and 3x-1=0.

$$x=5 x=\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$25x^{2}-40x+16-\left(4x+1\right)^{2}=0$$

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

$$25x^{2}-40x+16-\left(16x^{2}+8x+1\right)=0$$

To find the opposite of 16x^{2}+8x+1, find the opposite of each term.

$$25x^{2}-40x+16-16x^{2}-8x-1=0$$

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

$$9x^{2}-40x+16-8x-1=0$$

Combine -40x and -8x to get -48x.

$$9x^{2}-48x+16-1=0$$

Subtract 1 from 16 to get 15.

$$9x^{2}-48x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -48 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 9\times 15}}{2\times 9}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\times 9\times 15}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-48\right)±\sqrt{2304-36\times 15}}{2\times 9}$$

Multiply -36 times 15.

$$x=\frac{-\left(-48\right)±\sqrt{2304-540}}{2\times 9}$$

Add 2304 to -540.

$$x=\frac{-\left(-48\right)±\sqrt{1764}}{2\times 9}$$

Take the square root of 1764.

$$x=\frac{-\left(-48\right)±42}{2\times 9}$$

The opposite of -48 is 48.

$$x=\frac{48±42}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{48±42}{18}$$

Now solve the equation x=\frac{48±42}{18} when ± is plus. Add 48 to 42.

$$x=\frac{90}{18}$$

Divide 90 by 18.

$$x=5$$

Now solve the equation x=\frac{48±42}{18} when ± is minus. Subtract 42 from 48.

$$x=\frac{6}{18}$$

Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=5 x=\frac{1}{3}$$

Steps for Completing the Square

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

$$25x^{2}-40x+16-\left(4x+1\right)^{2}=0$$

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

$$25x^{2}-40x+16-\left(16x^{2}+8x+1\right)=0$$

To find the opposite of 16x^{2}+8x+1, find the opposite of each term.

$$25x^{2}-40x+16-16x^{2}-8x-1=0$$

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

$$9x^{2}-40x+16-8x-1=0$$

Combine -40x and -8x to get -48x.

$$9x^{2}-48x+16-1=0$$

Subtract 1 from 16 to get 15.

$$9x^{2}-48x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-48x=-15$$

Divide both sides by 9.

$$\frac{9x^{2}-48x}{9}=-\frac{15}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{48}{9}\right)x=-\frac{15}{9}$$

Reduce the fraction \frac{-48}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{16}{3}x=-\frac{15}{9}$$

Reduce the fraction \frac{-15}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{16}{3}x=-\frac{5}{3}$$

Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{3}x+\left(-\frac{8}{3}\right)^{2}=-\frac{5}{3}+\left(-\frac{8}{3}\right)^{2}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-\frac{5}{3}+\frac{64}{9}$$

Add -\frac{5}{3} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{16}{3}x+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{8}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{8}{3}=\frac{7}{3} x-\frac{8}{3}=-\frac{7}{3}$$

Add \frac{8}{3} to both sides of the equation.

$$x=5 x=\frac{1}{3}$$