Solve 2x^2-5x+2xsqrt{x^2-5x+6}=3

$$2 x ^ { 2 } - 5 x + 2 x \sqrt { x ^ { 2 } - 5 x + 6 } = 3$$

Solve for x

$$x=3$$

Solution Steps

Subtract 2x^{2}-5x from both sides of the equation.

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

To find the opposite of 2x^{2}-5x, find the opposite of each term.

$$2x\sqrt{x^{2}-5x+6}=3-2x^{2}+5x$$

Square both sides of the equation.

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

Expand \left(2x\sqrt{x^{2}-5x+6}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{x^{2}-5x+6} to the power of 2 and get x^{2}-5x+6.

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

Use the distributive property to multiply 4x^{2} by x^{2}-5x+6.

$$4x^{4}-20x^{3}+24x^{2}=\left(3-2x^{2}+5x\right)^{2}$$

Square 3-2x^{2}+5x.

$$4x^{4}-20x^{3}+24x^{2}=4x^{4}-20x^{3}+13x^{2}+30x+9$$

Subtract 4x^{4} from both sides.

$$4x^{4}-20x^{3}+24x^{2}-4x^{4}=-20x^{3}+13x^{2}+30x+9$$

Combine 4x^{4} and -4x^{4} to get 0.

$$-20x^{3}+24x^{2}=-20x^{3}+13x^{2}+30x+9$$

Add 20x^{3} to both sides.

$$-20x^{3}+24x^{2}+20x^{3}=13x^{2}+30x+9$$

Combine -20x^{3} and 20x^{3} to get 0.

$$24x^{2}=13x^{2}+30x+9$$

Subtract 13x^{2} from both sides.

$$24x^{2}-13x^{2}=30x+9$$

Combine 24x^{2} and -13x^{2} to get 11x^{2}.

$$11x^{2}=30x+9$$

Subtract 30x from both sides.

$$11x^{2}-30x=9$$

Subtract 9 from both sides.

$$11x^{2}-30x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 11x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-30 ab=11\left(-9\right)=-99$$

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. List all such integer pairs that give product -99.

$$1,-99 3,-33 9,-11$$

Calculate the sum for each pair.

$$1-99=-98 3-33=-30 9-11=-2$$

The solution is the pair that gives sum -30.

$$a=-33 b=3$$

Rewrite 11x^{2}-30x-9 as \left(11x^{2}-33x\right)+\left(3x-9\right).

$$\left(11x^{2}-33x\right)+\left(3x-9\right)$$

Factor out 11x in the first and 3 in the second group.

$$11x\left(x-3\right)+3\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(11x+3\right)$$

To find equation solutions, solve x-3=0 and 11x+3=0.

$$x=3 x=-\frac{3}{11}$$

Substitute 3 for x in the equation 2x^{2}-5x+2x\sqrt{x^{2}-5x+6}=3.

$$2\times 3^{2}-5\times 3+2\times 3\sqrt{3^{2}-5\times 3+6}=3$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Substitute -\frac{3}{11} for x in the equation 2x^{2}-5x+2x\sqrt{x^{2}-5x+6}=3.

$$2\left(-\frac{3}{11}\right)^{2}-5\left(-\frac{3}{11}\right)+2\left(-\frac{3}{11}\right)\sqrt{\left(-\frac{3}{11}\right)^{2}-5\left(-\frac{3}{11}\right)+6}=3$$

Simplify. The value x=-\frac{3}{11} does not satisfy the equation.

$$\frac{3}{121}=3$$

Equation 2x\sqrt{x^{2}-5x+6}=3+5x-2x^{2} has a unique solution.

$$x=3$$