Solve sqrt{5x^2+3x-1}-2x=1

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

Solve for x

$$x=2$$

Solution Steps

Subtract -2x from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{5x^{2}+3x-1}\right)^{2}=\left(1+2x\right)^{2}$$

Calculate \sqrt{5x^{2}+3x-1} to the power of 2 and get 5x^{2}+3x-1.

$$5x^{2}+3x-1=\left(1+2x\right)^{2}$$

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

$$5x^{2}+3x-1=1+4x+4x^{2}$$

Subtract 1 from both sides.

$$5x^{2}+3x-1-1=4x+4x^{2}$$

Subtract 1 from -1 to get -2.

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

Subtract 4x from both sides.

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

Combine 3x and -4x to get -x.

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

Subtract 4x^{2} from both sides.

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

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

To solve the equation, factor x^{2}-x-2 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.

$$a+b=-1 ab=-2$$

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

$$a=-2 b=1$$

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

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

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

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

Simplify. The value x=2 satisfies the equation.

$$1=1$$

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

$$\sqrt{5\left(-1\right)^{2}+3\left(-1\right)-1}-2\left(-1\right)=1$$

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

$$3=1$$

Equation \sqrt{5x^{2}+3x-1}=2x+1 has a unique solution.

$$x=2$$