Solve sqrt{x^3-7x+6}=x-1

$$\sqrt { x ^ { 3 } - 7 x + 6 } = x - 1$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{3}-7x+6}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{x^{3}-7x+6} to the power of 2 and get x^{3}-7x+6.

$$x^{3}-7x+6=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{3}-7x+6=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$x^{3}-7x+6-x^{2}=-2x+1$$

Add 2x to both sides.

$$x^{3}-7x+6-x^{2}+2x=1$$

Combine -7x and 2x to get -5x.

$$x^{3}-5x+6-x^{2}=1$$

Subtract 1 from both sides.

$$x^{3}-5x+6-x^{2}-1=0$$

Subtract 1 from 6 to get 5.

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-x^{2}-5x+5 by x-1 to get x^{2}-5. Solve the equation where the result equals to 0.

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

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, 0 for b, and -5 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{5}}{2}$$

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

$$x=-\sqrt{5} x=\sqrt{5}$$

List all found solutions.

$$x=1 x=-\sqrt{5} x=\sqrt{5}$$

Substitute 1 for x in the equation \sqrt{x^{3}-7x+6}=x-1.

$$\sqrt{1^{3}-7+6}=1-1$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -\sqrt{5} for x in the equation \sqrt{x^{3}-7x+6}=x-1.

$$\sqrt{\left(-\sqrt{5}\right)^{3}-7\left(-\sqrt{5}\right)+6}=-\sqrt{5}-1$$

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

$$5^{\frac{1}{2}}+1=-5^{\frac{1}{2}}-1$$

Substitute \sqrt{5} for x in the equation \sqrt{x^{3}-7x+6}=x-1.

$$\sqrt{\left(\sqrt{5}\right)^{3}-7\sqrt{5}+6}=\sqrt{5}-1$$

Simplify. The value x=\sqrt{5} satisfies the equation.

$$5^{\frac{1}{2}}-1=5^{\frac{1}{2}}-1$$

List all solutions of \sqrt{x^{3}-7x+6}=x-1.

$$x=1 x=\sqrt{5}$$