Solve 3x+5=x^2+1

$$3 x + 5 = x ^ { 2 } + 1$$

Solve for x

$$x=-1$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$3x+5-x^{2}=1$$

Subtract 1 from both sides.

$$3x+5-x^{2}-1=0$$

Subtract 1 from 5 to get 4.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=3 ab=-4=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).

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

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

$$-x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-1\right)$$

To find equation solutions, solve x-4=0 and -x-1=0.

$$x=4 x=-1$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$3x+5-x^{2}=1$$

Subtract 1 from both sides.

$$3x+5-x^{2}-1=0$$

Subtract 1 from 5 to get 4.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

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

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±5}{-2}$$

Now solve the equation x=\frac{-3±5}{-2} when ± is plus. Add -3 to 5.

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

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-3±5}{-2} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-1 x=4$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$3x+5-x^{2}=1$$

Subtract 5 from both sides.

$$3x-x^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$3x-x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

$$x^{2}-3x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x^{2}-3x=4$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=4 x=-1$$