Solve left(x-1right)^2-4(x^2-1)=0

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

Solve for x

$$x=1$$

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply -4 by x^{2}-1.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-2x+1+4=0$$

Add 1 and 4 to get 5.

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

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.

$$a+b=-2 ab=-3\times 5=-15$$

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 -15.

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

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=3 b=-5$$

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

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

Factor out 3x in the first and 5 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply -4 by x^{2}-1.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-2x+1+4=0$$

Add 1 and 4 to get 5.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\times 5}}{2\left(-3\right)}$$

Multiply 12 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\left(-3\right)}$$

Add 4 to 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-\left(-2\right)±8}{2\left(-3\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{2±8}{-6}$$

Now solve the equation x=\frac{2±8}{-6} when ± is plus. Add 2 to 8.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{2±8}{-6} when ± is minus. Subtract 8 from 2.

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

Divide -6 by -6.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Use the distributive property to multiply -4 by x^{2}-1.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-2x+1+4=0$$

Add 1 and 4 to get 5.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -2 by -3.

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

Divide -5 by -3.

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

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{5}{3}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{3}+\frac{1}{9}$$

Add \frac{5}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{16}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{4}{3} x+\frac{1}{3}=-\frac{4}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

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