Solve x(x^2-9)+10=0

$$x ( x ^ { 2 } - 9 ) + 10 = 0$$

Solve for x (complex solution)

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\left(\sqrt{6}+1\right)\approx -3.449489743$$

$$x=2$$

Solution Steps

Multiply and combine like terms.

$$x^{3}-9x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x+10 by x-2 to get x^{2}+2x-5. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 2 for b, and -5 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Solve the equation x^{2}+2x-5=0 when ± is plus and when ± is minus.

$$x=-\sqrt{6}-1 x=\sqrt{6}-1$$

List all found solutions.

$$x=2 x=-\sqrt{6}-1 x=\sqrt{6}-1$$

Solve for x

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\sqrt{6}-1\approx -3.449489743$$

$$x=2$$

Solution Steps

Multiply and combine like terms.

$$x^{3}-9x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x+10 by x-2 to get x^{2}+2x-5. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 2 for b, and -5 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Solve the equation x^{2}+2x-5=0 when ± is plus and when ± is minus.

$$x=-\sqrt{6}-1 x=\sqrt{6}-1$$

List all found solutions.

$$x=2 x=-\sqrt{6}-1 x=\sqrt{6}-1$$