Solve (x^2-4)sqrt{x-1}=0

$$( { x }^{ 2 } -4) \sqrt{ x-1 } =0$$

Solve for x

$$x=1$$

$$x=2$$

Solution Steps

Use the distributive property to multiply x^{2}-4 by \sqrt{x-1}.

$$x^{2}\sqrt{x-1}-4\sqrt{x-1}=0$$

Subtract -4\sqrt{x-1} from both sides of the equation.

$$x^{2}\sqrt{x-1}=4\sqrt{x-1}$$

Square both sides of the equation.

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

Expand \left(x^{2}\sqrt{x-1}\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Calculate \sqrt{x-1} to the power of 2 and get x-1.

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

Use the distributive property to multiply x^{4} by x-1.

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

Expand \left(4\sqrt{x-1}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$x^{5}-x^{4}=16\left(\sqrt{x-1}\right)^{2}$$

Calculate \sqrt{x-1} to the power of 2 and get x-1.

$$x^{5}-x^{4}=16\left(x-1\right)$$

Use the distributive property to multiply 16 by x-1.

$$x^{5}-x^{4}=16x-16$$

Subtract 16x from both sides.

$$x^{5}-x^{4}-16x=-16$$

Add 16 to both sides.

$$x^{5}-x^{4}-16x+16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}-x^{4}-16x+16 by x-1 to get x^{4}-16. Solve the equation where the result equals to 0.

$$x^{4}-16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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^{4}-16 by x-2 to get x^{3}+2x^{2}+4x+8. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}+4x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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}+2x^{2}+4x+8 by x+2 to get x^{2}+4. Solve the equation where the result equals to 0.

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

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 4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-16}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1 x=2 x=-2$$

Substitute 1 for x in the equation \left(x^{2}-4\right)\sqrt{x-1}=0.

$$\left(1^{2}-4\right)\sqrt{1-1}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute 2 for x in the equation \left(x^{2}-4\right)\sqrt{x-1}=0.

$$\left(2^{2}-4\right)\sqrt{2-1}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation \left(x^{2}-4\right)\sqrt{x-1}=0. The expression \sqrt{-2-1} is undefined because the radicand cannot be negative.

$$\left(\left(-2\right)^{2}-4\right)\sqrt{-2-1}=0$$

List all solutions of \sqrt{x-1}x^{2}=4\sqrt{x-1}.

$$x=1 x=2$$

Solve for x (complex solution)

$$x=2$$

$$x=-2$$

$$x=1$$