Solve 2x-5sqrt{x}+2=0

$$2 x - 5 \sqrt { x } + 2 = 0$$

Solve for x

$$x=4$$

$$x=\frac{1}{4}=0.25$$

Solution Steps

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Subtract 2x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Calculate -5 to the power of 2 and get 25.

$$25\left(\sqrt{x}\right)^{2}=\left(-2-2x\right)^{2}$$

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

$$25x=\left(-2-2x\right)^{2}$$

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

$$25x=4+8x+4x^{2}$$

Subtract 8x from both sides.

$$25x-8x=4+4x^{2}$$

Combine 25x and -8x to get 17x.

$$17x=4+4x^{2}$$

Subtract 4x^{2} from both sides.

$$17x-4x^{2}=4$$

Subtract 4 from both sides.

$$17x-4x^{2}-4=0$$

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

$$-4x^{2}+17x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=17 ab=-4\left(-4\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 17.

$$a=16 b=1$$

Rewrite -4x^{2}+17x-4 as \left(-4x^{2}+16x\right)+\left(x-4\right).

$$\left(-4x^{2}+16x\right)+\left(x-4\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(-x+4\right)-\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(4x-1\right)$$

To find equation solutions, solve -x+4=0 and 4x-1=0.

$$x=4 x=\frac{1}{4}$$

Substitute 4 for x in the equation 2x-5\sqrt{x}+2=0.

$$2\times 4-5\sqrt{4}+2=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute \frac{1}{4} for x in the equation 2x-5\sqrt{x}+2=0.

$$2\times \frac{1}{4}-5\sqrt{\frac{1}{4}}+2=0$$

Simplify. The value x=\frac{1}{4} satisfies the equation.

$$0=0$$

List all solutions of -5\sqrt{x}=-2x-2.

$$x=4 x=\frac{1}{4}$$