Solve (4)4(x-1)^2-9(x-5)^2=0

$$( 4 ) 4 ( x - 1 ) ^ { 2 } - 9 ( x - 5 ) ^ { 2 } = 0$$

Solve for x

$$x=-11$$

$$x = \frac{19}{7} = 2\frac{5}{7} \approx 2.714285714$$

Steps Using Factoring By Grouping

Multiply 4 and 4 to get 16.

$$16\left(x-1\right)^{2}-9\left(x-5\right)^{2}=0$$

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

$$16\left(x^{2}-2x+1\right)-9\left(x-5\right)^{2}=0$$

Use the distributive property to multiply 16 by x^{2}-2x+1.

$$16x^{2}-32x+16-9\left(x-5\right)^{2}=0$$

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

$$16x^{2}-32x+16-9\left(x^{2}-10x+25\right)=0$$

Use the distributive property to multiply -9 by x^{2}-10x+25.

$$16x^{2}-32x+16-9x^{2}+90x-225=0$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-32x+16+90x-225=0$$

Combine -32x and 90x to get 58x.

$$7x^{2}+58x+16-225=0$$

Subtract 225 from 16 to get -209.

$$7x^{2}+58x-209=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-209. To find a and b, set up a system to be solved.

$$a+b=58 ab=7\left(-209\right)=-1463$$

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. List all such integer pairs that give product -1463.

$$-1,1463 -7,209 -11,133 -19,77$$

Calculate the sum for each pair.

$$-1+1463=1462 -7+209=202 -11+133=122 -19+77=58$$

The solution is the pair that gives sum 58.

$$a=-19 b=77$$

Rewrite 7x^{2}+58x-209 as \left(7x^{2}-19x\right)+\left(77x-209\right).

$$\left(7x^{2}-19x\right)+\left(77x-209\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(7x-19\right)+11\left(7x-19\right)$$

Factor out common term 7x-19 by using distributive property.

$$\left(7x-19\right)\left(x+11\right)$$

To find equation solutions, solve 7x-19=0 and x+11=0.

$$x=\frac{19}{7} x=-11$$

Steps Using the Quadratic Formula

Multiply 4 and 4 to get 16.

$$16\left(x-1\right)^{2}-9\left(x-5\right)^{2}=0$$

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

$$16\left(x^{2}-2x+1\right)-9\left(x-5\right)^{2}=0$$

Use the distributive property to multiply 16 by x^{2}-2x+1.

$$16x^{2}-32x+16-9\left(x-5\right)^{2}=0$$

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

$$16x^{2}-32x+16-9\left(x^{2}-10x+25\right)=0$$

Use the distributive property to multiply -9 by x^{2}-10x+25.

$$16x^{2}-32x+16-9x^{2}+90x-225=0$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-32x+16+90x-225=0$$

Combine -32x and 90x to get 58x.

$$7x^{2}+58x+16-225=0$$

Subtract 225 from 16 to get -209.

$$7x^{2}+58x-209=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 58 for b, and -209 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-58±\sqrt{58^{2}-4\times 7\left(-209\right)}}{2\times 7}$$

Square 58.

$$x=\frac{-58±\sqrt{3364-4\times 7\left(-209\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-58±\sqrt{3364-28\left(-209\right)}}{2\times 7}$$

Multiply -28 times -209.

$$x=\frac{-58±\sqrt{3364+5852}}{2\times 7}$$

Add 3364 to 5852.

$$x=\frac{-58±\sqrt{9216}}{2\times 7}$$

Take the square root of 9216.

$$x=\frac{-58±96}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-58±96}{14}$$

Now solve the equation x=\frac{-58±96}{14} when ± is plus. Add -58 to 96.

$$x=\frac{38}{14}$$

Reduce the fraction \frac{38}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{19}{7}$$

Now solve the equation x=\frac{-58±96}{14} when ± is minus. Subtract 96 from -58.

$$x=-\frac{154}{14}$$

Divide -154 by 14.

$$x=-11$$

The equation is now solved.

$$x=\frac{19}{7} x=-11$$

Steps for Completing the Square

Multiply 4 and 4 to get 16.

$$16\left(x-1\right)^{2}-9\left(x-5\right)^{2}=0$$

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

$$16\left(x^{2}-2x+1\right)-9\left(x-5\right)^{2}=0$$

Use the distributive property to multiply 16 by x^{2}-2x+1.

$$16x^{2}-32x+16-9\left(x-5\right)^{2}=0$$

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

$$16x^{2}-32x+16-9\left(x^{2}-10x+25\right)=0$$

Use the distributive property to multiply -9 by x^{2}-10x+25.

$$16x^{2}-32x+16-9x^{2}+90x-225=0$$

Combine 16x^{2} and -9x^{2} to get 7x^{2}.

$$7x^{2}-32x+16+90x-225=0$$

Combine -32x and 90x to get 58x.

$$7x^{2}+58x+16-225=0$$

Subtract 225 from 16 to get -209.

$$7x^{2}+58x-209=0$$

Add 209 to both sides. Anything plus zero gives itself.

$$7x^{2}+58x=209$$

Divide both sides by 7.

$$\frac{7x^{2}+58x}{7}=\frac{209}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{58}{7}x=\frac{209}{7}$$

Divide \frac{58}{7}, the coefficient of the x term, by 2 to get \frac{29}{7}. Then add the square of \frac{29}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{58}{7}x+\left(\frac{29}{7}\right)^{2}=\frac{209}{7}+\left(\frac{29}{7}\right)^{2}$$

Square \frac{29}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{58}{7}x+\frac{841}{49}=\frac{209}{7}+\frac{841}{49}$$

Add \frac{209}{7} to \frac{841}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{58}{7}x+\frac{841}{49}=\frac{2304}{49}$$

Factor x^{2}+\frac{58}{7}x+\frac{841}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{29}{7}\right)^{2}=\frac{2304}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{29}{7}\right)^{2}}=\sqrt{\frac{2304}{49}}$$

Simplify.

$$x+\frac{29}{7}=\frac{48}{7} x+\frac{29}{7}=-\frac{48}{7}$$

Subtract \frac{29}{7} from both sides of the equation.

$$x=\frac{19}{7} x=-11$$