Solve sqrt{x^3-4x^2-10x+29}=3-x

$$\sqrt { x ^ { 3 } - 4 x ^ { 2 } - 10 x + 29 } = 3 - x$$

Solve for x

$$x=2$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{3}-4x^{2}-10x+29}\right)^{2}=\left(3-x\right)^{2}$$

Calculate \sqrt{x^{3}-4x^{2}-10x+29} to the power of 2 and get x^{3}-4x^{2}-10x+29.

$$x^{3}-4x^{2}-10x+29=\left(3-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$x^{3}-4x^{2}-10x+29=9-6x+x^{2}$$

Subtract 9 from both sides.

$$x^{3}-4x^{2}-10x+29-9=-6x+x^{2}$$

Subtract 9 from 29 to get 20.

$$x^{3}-4x^{2}-10x+20=-6x+x^{2}$$

Add 6x to both sides.

$$x^{3}-4x^{2}-10x+20+6x=x^{2}$$

Combine -10x and 6x to get -4x.

$$x^{3}-4x^{2}-4x+20=x^{2}$$

Subtract x^{2} from both sides.

$$x^{3}-4x^{2}-4x+20-x^{2}=0$$

Combine -4x^{2} and -x^{2} to get -5x^{2}.

$$x^{3}-5x^{2}-4x+20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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}-5x^{2}-4x+20 by x-2 to get x^{2}-3x-10. Solve the equation where the result equals to 0.

$$x^{2}-3x-10=0$$

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, -3 for b, and -10 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{3±7}{2}$$

Solve the equation x^{2}-3x-10=0 when ± is plus and when ± is minus.

$$x=-2 x=5$$

List all found solutions.

$$x=2 x=-2 x=5$$

Substitute 2 for x in the equation \sqrt{x^{3}-4x^{2}-10x+29}=3-x.

$$\sqrt{2^{3}-4\times 2^{2}-10\times 2+29}=3-2$$

Simplify. The value x=2 satisfies the equation.

$$1=1$$

Substitute -2 for x in the equation \sqrt{x^{3}-4x^{2}-10x+29}=3-x.

$$\sqrt{\left(-2\right)^{3}-4\left(-2\right)^{2}-10\left(-2\right)+29}=3-\left(-2\right)$$

Simplify. The value x=-2 satisfies the equation.

$$5=5$$

Substitute 5 for x in the equation \sqrt{x^{3}-4x^{2}-10x+29}=3-x.

$$\sqrt{5^{3}-4\times 5^{2}-10\times 5+29}=3-5$$

Simplify. The value x=5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

List all solutions of \sqrt{x^{3}-4x^{2}-10x+29}=3-x.

$$x=2 x=-2$$