Solve x^3-7x^2+5=x(x^2-1)+3x^2-2

$$x ^ { 3 } - 7 x ^ { 2 } + 5 = x ( x ^ { 2 } - 1 ) + 3 x ^ { 2 } - 2$$

Solve for x

$$x=\frac{\sqrt{281}+1}{20}\approx 0.888152731$$

$$x=\frac{1-\sqrt{281}}{20}\approx -0.788152731$$

Steps Using the Quadratic Formula

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

$$x^{3}-7x^{2}+5=x^{3}-x+3x^{2}-2$$

Subtract x^{3} from both sides.

$$x^{3}-7x^{2}+5-x^{3}=-x+3x^{2}-2$$

Combine x^{3} and -x^{3} to get 0.

$$-7x^{2}+5=-x+3x^{2}-2$$

Add x to both sides.

$$-7x^{2}+5+x=3x^{2}-2$$

Subtract 3x^{2} from both sides.

$$-7x^{2}+5+x-3x^{2}=-2$$

Combine -7x^{2} and -3x^{2} to get -10x^{2}.

$$-10x^{2}+5+x=-2$$

Add 2 to both sides.

$$-10x^{2}+5+x+2=0$$

Add 5 and 2 to get 7.

$$-10x^{2}+7+x=0$$

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.

$$-10x^{2}+x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 1 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-10\right)\times 7}}{2\left(-10\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-10\right)\times 7}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-1±\sqrt{1+40\times 7}}{2\left(-10\right)}$$

Multiply 40 times 7.

$$x=\frac{-1±\sqrt{1+280}}{2\left(-10\right)}$$

Add 1 to 280.

$$x=\frac{-1±\sqrt{281}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-1±\sqrt{281}}{-20}$$

Now solve the equation x=\frac{-1±\sqrt{281}}{-20} when ± is plus. Add -1 to \sqrt{281}.

$$x=\frac{\sqrt{281}-1}{-20}$$

Divide -1+\sqrt{281} by -20.

$$x=\frac{1-\sqrt{281}}{20}$$

Now solve the equation x=\frac{-1±\sqrt{281}}{-20} when ± is minus. Subtract \sqrt{281} from -1.

$$x=\frac{-\sqrt{281}-1}{-20}$$

Divide -1-\sqrt{281} by -20.

$$x=\frac{\sqrt{281}+1}{20}$$

The equation is now solved.

$$x=\frac{1-\sqrt{281}}{20} x=\frac{\sqrt{281}+1}{20}$$

Steps for Completing the Square

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

$$x^{3}-7x^{2}+5=x^{3}-x+3x^{2}-2$$

Subtract x^{3} from both sides.

$$x^{3}-7x^{2}+5-x^{3}=-x+3x^{2}-2$$

Combine x^{3} and -x^{3} to get 0.

$$-7x^{2}+5=-x+3x^{2}-2$$

Add x to both sides.

$$-7x^{2}+5+x=3x^{2}-2$$

Subtract 3x^{2} from both sides.

$$-7x^{2}+5+x-3x^{2}=-2$$

Combine -7x^{2} and -3x^{2} to get -10x^{2}.

$$-10x^{2}+5+x=-2$$

Subtract 5 from both sides.

$$-10x^{2}+x=-2-5$$

Subtract 5 from -2 to get -7.

$$-10x^{2}+x=-7$$

Divide both sides by -10.

$$\frac{-10x^{2}+x}{-10}=-\frac{7}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{1}{-10}x=-\frac{7}{-10}$$

Divide 1 by -10.

$$x^{2}-\frac{1}{10}x=-\frac{7}{-10}$$

Divide -7 by -10.

$$x^{2}-\frac{1}{10}x=\frac{7}{10}$$

Divide -\frac{1}{10}, the coefficient of the x term, by 2 to get -\frac{1}{20}. Then add the square of -\frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\frac{7}{10}+\left(-\frac{1}{20}\right)^{2}$$

Square -\frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{7}{10}+\frac{1}{400}$$

Add \frac{7}{10} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{281}{400}$$

Factor x^{2}-\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{281}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{281}{400}}$$

Simplify.

$$x-\frac{1}{20}=\frac{\sqrt{281}}{20} x-\frac{1}{20}=-\frac{\sqrt{281}}{20}$$

Add \frac{1}{20} to both sides of the equation.

$$x=\frac{\sqrt{281}+1}{20} x=\frac{1-\sqrt{281}}{20}$$