Solve (5x-4)^2-49x^2=0

$$( 5 x - 4 ) ^ { 2 } - 49 x ^ { 2 } = 0$$

Solve for x

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

$$x=-2$$

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}.

$$25x^{2}-40x+16-49x^{2}=0$$

Combine 25x^{2} and -49x^{2} to get -24x^{2}.

$$-24x^{2}-40x+16=0$$

Divide both sides by 8.

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

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

$$a+b=-5 ab=-3\times 2=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -3x^{2}-5x+2 as \left(-3x^{2}+x\right)+\left(-6x+2\right).

$$\left(-3x^{2}+x\right)+\left(-6x+2\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(3x-1\right)-2\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-x-2\right)$$

To find equation solutions, solve 3x-1=0 and -x-2=0.

$$x=\frac{1}{3} x=-2$$

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}.

$$25x^{2}-40x+16-49x^{2}=0$$

Combine 25x^{2} and -49x^{2} to get -24x^{2}.

$$-24x^{2}-40x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -24 for a, -40 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-24\right)\times 16}}{2\left(-24\right)}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\left(-24\right)\times 16}}{2\left(-24\right)}$$

Multiply -4 times -24.

$$x=\frac{-\left(-40\right)±\sqrt{1600+96\times 16}}{2\left(-24\right)}$$

Multiply 96 times 16.

$$x=\frac{-\left(-40\right)±\sqrt{1600+1536}}{2\left(-24\right)}$$

Add 1600 to 1536.

$$x=\frac{-\left(-40\right)±\sqrt{3136}}{2\left(-24\right)}$$

Take the square root of 3136.

$$x=\frac{-\left(-40\right)±56}{2\left(-24\right)}$$

The opposite of -40 is 40.

$$x=\frac{40±56}{2\left(-24\right)}$$

Multiply 2 times -24.

$$x=\frac{40±56}{-48}$$

Now solve the equation x=\frac{40±56}{-48} when ± is plus. Add 40 to 56.

$$x=\frac{96}{-48}$$

Divide 96 by -48.

$$x=-2$$

Now solve the equation x=\frac{40±56}{-48} when ± is minus. Subtract 56 from 40.

$$x=-\frac{16}{-48}$$

Reduce the fraction \frac{-16}{-48} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

$$x=-2 x=\frac{1}{3}$$

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}.

$$25x^{2}-40x+16-49x^{2}=0$$

Combine 25x^{2} and -49x^{2} to get -24x^{2}.

$$-24x^{2}-40x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-24x^{2}-40x=-16$$

Divide both sides by -24.

$$\frac{-24x^{2}-40x}{-24}=-\frac{16}{-24}$$

Dividing by -24 undoes the multiplication by -24.

$$x^{2}+\left(-\frac{40}{-24}\right)x=-\frac{16}{-24}$$

Reduce the fraction \frac{-40}{-24} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{5}{3}x=-\frac{16}{-24}$$

Reduce the fraction \frac{-16}{-24} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{5}{3}x=\frac{2}{3}$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=\frac{2}{3}+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{2}{3}+\frac{25}{36}$$

Add \frac{2}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{49}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{7}{6} x+\frac{5}{6}=-\frac{7}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$x=\frac{1}{3} x=-2$$