Solve sqrt{x^2+5x-1}=sqrt{x+4}

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

Solve for x (complex solution)

$$x=-5$$

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+5x-1} to the power of 2 and get x^{2}+5x-1.

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

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

$$x^{2}+5x-1=x+4$$

Subtract x from both sides.

$$x^{2}+5x-1-x=4$$

Combine 5x and -x to get 4x.

$$x^{2}+4x-1=4$$

Subtract 4 from both sides.

$$x^{2}+4x-1-4=0$$

Subtract 4 from -1 to get -5.

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

To solve the equation, factor x^{2}+4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve x-1=0 and x+5=0.

$$x=1 x=-5$$

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

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

Simplify. The value x=1 satisfies the equation.

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

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

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

Simplify. The value x=-5 satisfies the equation.

$$i=i$$

List all solutions of \sqrt{x^{2}+5x-1}=\sqrt{x+4}.

$$x=1 x=-5$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+5x-1} to the power of 2 and get x^{2}+5x-1.

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

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

$$x^{2}+5x-1=x+4$$

Subtract x from both sides.

$$x^{2}+5x-1-x=4$$

Combine 5x and -x to get 4x.

$$x^{2}+4x-1=4$$

Subtract 4 from both sides.

$$x^{2}+4x-1-4=0$$

Subtract 4 from -1 to get -5.

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

To solve the equation, factor x^{2}+4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve x-1=0 and x+5=0.

$$x=1 x=-5$$

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

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

Simplify. The value x=1 satisfies the equation.

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

Substitute -5 for x in the equation \sqrt{x^{2}+5x-1}=\sqrt{x+4}. The expression \sqrt{\left(-5\right)^{2}+5\left(-5\right)-1} is undefined because the radicand cannot be negative.

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

Equation \sqrt{x^{2}+5x-1}=\sqrt{x+4} has a unique solution.

$$x=1$$