Solve (a^2-1)2+a^2+3a-4=0

$$( a ^ { 2 } - 1 ) 2 + a ^ { 2 } + 3 a - 4 = 0$$

Solve for a

$$a=-2$$

$$a=1$$

Steps Using Factoring By Grouping

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

$$2a^{2}-2+a^{2}+3a-4=0$$

Combine 2a^{2} and a^{2} to get 3a^{2}.

$$3a^{2}-2+3a-4=0$$

Subtract 4 from -2 to get -6.

$$3a^{2}-6+3a=0$$

Divide both sides by 3.

$$a^{2}-2+a=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{2}+a-2=0$$

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

$$a+b=1 ab=1\left(-2\right)=-2$$

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. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite a^{2}+a-2 as \left(a^{2}-a\right)+\left(2a-2\right).

$$\left(a^{2}-a\right)+\left(2a-2\right)$$

Factor out a in the first and 2 in the second group.

$$a\left(a-1\right)+2\left(a-1\right)$$

Factor out common term a-1 by using distributive property.

$$\left(a-1\right)\left(a+2\right)$$

To find equation solutions, solve a-1=0 and a+2=0.

$$a=1 a=-2$$

Steps Using the Quadratic Formula

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

$$2a^{2}-2+a^{2}+3a-4=0$$

Combine 2a^{2} and a^{2} to get 3a^{2}.

$$3a^{2}-2+3a-4=0$$

Subtract 4 from -2 to get -6.

$$3a^{2}-6+3a=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3a^{2}+3a-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-3±\sqrt{3^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Square 3.

$$a=\frac{-3±\sqrt{9-4\times 3\left(-6\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-3±\sqrt{9-12\left(-6\right)}}{2\times 3}$$

Multiply -12 times -6.

$$a=\frac{-3±\sqrt{9+72}}{2\times 3}$$

Add 9 to 72.

$$a=\frac{-3±\sqrt{81}}{2\times 3}$$

Take the square root of 81.

$$a=\frac{-3±9}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{-3±9}{6}$$

Now solve the equation a=\frac{-3±9}{6} when ± is plus. Add -3 to 9.

$$a=\frac{6}{6}$$

Divide 6 by 6.

$$a=1$$

Now solve the equation a=\frac{-3±9}{6} when ± is minus. Subtract 9 from -3.

$$a=-\frac{12}{6}$$

Divide -12 by 6.

$$a=-2$$

The equation is now solved.

$$a=1 a=-2$$

Steps for Completing the Square

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

$$2a^{2}-2+a^{2}+3a-4=0$$

Combine 2a^{2} and a^{2} to get 3a^{2}.

$$3a^{2}-2+3a-4=0$$

Subtract 4 from -2 to get -6.

$$3a^{2}-6+3a=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$3a^{2}+3a=6$$

Divide both sides by 3.

$$\frac{3a^{2}+3a}{3}=\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a^{2}+\frac{3}{3}a=\frac{6}{3}$$

Divide 3 by 3.

$$a^{2}+a=\frac{6}{3}$$

Divide 6 by 3.

$$a^{2}+a=2$$

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

$$a^{2}+a+\left(\frac{1}{2}\right)^{2}=2+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+a+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$a^{2}+a+\frac{1}{4}=\frac{9}{4}$$

Factor a^{2}+a+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$a+\frac{1}{2}=\frac{3}{2} a+\frac{1}{2}=-\frac{3}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$a=1 a=-2$$