Solve x^2+3x-4=5sqrt{x^2+3x-4}

$$x ^ { 2 } + 3 x - 4 = 5 \sqrt { x ^ { 2 } + 3 x - 4 }$$

Solve for x

$$x=\frac{-5\sqrt{5}-3}{2}\approx -7.090169944$$

$$x = \frac{5 \sqrt{5} - 3}{2} \approx 4.090169944$$

$$x=1$$

$$x=-4$$

Solution Steps

Square both sides of the equation.

$$\left(x^{2}+3x-4\right)^{2}=\left(5\sqrt{x^{2}+3x-4}\right)^{2}$$

Square x^{2}+3x-4.

$$x^{4}+6x^{3}+x^{2}-24x+16=\left(5\sqrt{x^{2}+3x-4}\right)^{2}$$

Expand \left(5\sqrt{x^{2}+3x-4}\right)^{2}.

$$x^{4}+6x^{3}+x^{2}-24x+16=5^{2}\left(\sqrt{x^{2}+3x-4}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$x^{4}+6x^{3}+x^{2}-24x+16=25\left(\sqrt{x^{2}+3x-4}\right)^{2}$$

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

$$x^{4}+6x^{3}+x^{2}-24x+16=25\left(x^{2}+3x-4\right)$$

Use the distributive property to multiply 25 by x^{2}+3x-4.

$$x^{4}+6x^{3}+x^{2}-24x+16=25x^{2}+75x-100$$

Subtract 25x^{2} from both sides.

$$x^{4}+6x^{3}+x^{2}-24x+16-25x^{2}=75x-100$$

Combine x^{2} and -25x^{2} to get -24x^{2}.

$$x^{4}+6x^{3}-24x^{2}-24x+16=75x-100$$

Subtract 75x from both sides.

$$x^{4}+6x^{3}-24x^{2}-24x+16-75x=-100$$

Combine -24x and -75x to get -99x.

$$x^{4}+6x^{3}-24x^{2}-99x+16=-100$$

Add 100 to both sides.

$$x^{4}+6x^{3}-24x^{2}-99x+16+100=0$$

Add 16 and 100 to get 116.

$$x^{4}+6x^{3}-24x^{2}-99x+116=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 116 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±116,±58,±29,±4,±2,±1$$

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^{4}+6x^{3}-24x^{2}-99x+116 by x-1 to get x^{3}+7x^{2}-17x-116. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}-17x-116=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -116 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±116,±58,±29,±4,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+7x^{2}-17x-116 by x+4 to get x^{2}+3x-29. Solve the equation where the result equals to 0.

$$x^{2}+3x-29=0$$

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 -29 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-29\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±5\sqrt{5}}{2}$$

Solve the equation x^{2}+3x-29=0 when ± is plus and when ± is minus.

$$x=\frac{-5\sqrt{5}-3}{2} x=\frac{5\sqrt{5}-3}{2}$$

List all found solutions.

$$x=1 x=-4 x=\frac{-5\sqrt{5}-3}{2} x=\frac{5\sqrt{5}-3}{2}$$

Substitute 1 for x in the equation x^{2}+3x-4=5\sqrt{x^{2}+3x-4}.

$$1^{2}+3\times 1-4=5\sqrt{1^{2}+3\times 1-4}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

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

$$\left(-4\right)^{2}+3\left(-4\right)-4=5\sqrt{\left(-4\right)^{2}+3\left(-4\right)-4}$$

Simplify. The value x=-4 satisfies the equation.

$$0=0$$

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

$$\left(\frac{-5\sqrt{5}-3}{2}\right)^{2}+3\times \frac{-5\sqrt{5}-3}{2}-4=5\sqrt{\left(\frac{-5\sqrt{5}-3}{2}\right)^{2}+3\times \frac{-5\sqrt{5}-3}{2}-4}$$

Simplify. The value x=\frac{-5\sqrt{5}-3}{2} satisfies the equation.

$$25=25$$

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

$$\left(\frac{5\sqrt{5}-3}{2}\right)^{2}+3\times \frac{5\sqrt{5}-3}{2}-4=5\sqrt{\left(\frac{5\sqrt{5}-3}{2}\right)^{2}+3\times \frac{5\sqrt{5}-3}{2}-4}$$

Simplify. The value x=\frac{5\sqrt{5}-3}{2} satisfies the equation.

$$25=25$$

List all solutions of x^{2}+3x-4=5\sqrt{x^{2}+3x-4}.

$$x=1 x=-4 x=\frac{-5\sqrt{5}-3}{2} x=\frac{5\sqrt{5}-3}{2}$$