Solve x^2-1/5=sqrt{x}+1

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

Solve for x

$$x=4$$

Solution Steps

Subtract 1 from both sides of the equation.

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

Multiply both sides of the equation by 5.

$$x^{2}-1-5=5\sqrt{x}$$

Subtract 5 from -1 to get -6.

$$x^{2}-6=5\sqrt{x}$$

Square both sides of the equation.

$$\left(x^{2}-6\right)^{2}=\left(5\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-6\right)^{2}.

$$\left(x^{2}\right)^{2}-12x^{2}+36=\left(5\sqrt{x}\right)^{2}$$

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

$$x^{4}-12x^{2}+36=\left(5\sqrt{x}\right)^{2}$$

Expand \left(5\sqrt{x}\right)^{2}.

$$x^{4}-12x^{2}+36=5^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$x^{4}-12x^{2}+36=25\left(\sqrt{x}\right)^{2}$$

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

$$x^{4}-12x^{2}+36=25x$$

Subtract 25x from both sides.

$$x^{4}-12x^{2}+36-25x=0$$

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

$$x^{4}-12x^{2}-25x+36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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^{4}-12x^{2}-25x+36 by x-1 to get x^{3}+x^{2}-11x-36. Solve the equation where the result equals to 0.

$$x^{3}+x^{2}-11x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+x^{2}-11x-36 by x-4 to get x^{2}+5x+9. Solve the equation where the result equals to 0.

$$x^{2}+5x+9=0$$

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, 5 for b, and 9 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-11}}{2}$$

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.

$$x=1 x=4$$

Substitute 1 for x in the equation \frac{x^{2}-1}{5}=\sqrt{x}+1.

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

Simplify. The value x=1 does not satisfy the equation.

$$0=2$$

Substitute 4 for x in the equation \frac{x^{2}-1}{5}=\sqrt{x}+1.

$$\frac{4^{2}-1}{5}=\sqrt{4}+1$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation x^{2}-6=5\sqrt{x} has a unique solution.

$$x=4$$