Solve 2x^2-10x=6+11sqrt{x^2-5x}

$$2 x ^ { 2 } - 10 x = 6 + 11 \sqrt { x ^ { 2 } - 5 x }$$

Solve for x

$$x=-4$$

$$x=9$$

Solution Steps

Subtract 6 from both sides of the equation.

$$2x^{2}-10x-6=11\sqrt{x^{2}-5x}$$

Square both sides of the equation.

$$\left(2x^{2}-10x-6\right)^{2}=\left(11\sqrt{x^{2}-5x}\right)^{2}$$

Square 2x^{2}-10x-6.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=\left(11\sqrt{x^{2}-5x}\right)^{2}$$

Expand \left(11\sqrt{x^{2}-5x}\right)^{2}.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=11^{2}\left(\sqrt{x^{2}-5x}\right)^{2}$$

Calculate 11 to the power of 2 and get 121.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121\left(\sqrt{x^{2}-5x}\right)^{2}$$

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

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121\left(x^{2}-5x\right)$$

Use the distributive property to multiply 121 by x^{2}-5x.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121x^{2}-605x$$

Subtract 121x^{2} from both sides.

$$4x^{4}-40x^{3}+76x^{2}+120x+36-121x^{2}=-605x$$

Combine 76x^{2} and -121x^{2} to get -45x^{2}.

$$4x^{4}-40x^{3}-45x^{2}+120x+36=-605x$$

Add 605x to both sides.

$$4x^{4}-40x^{3}-45x^{2}+120x+36+605x=0$$

Combine 120x and 605x to get 725x.

$$4x^{4}-40x^{3}-45x^{2}+725x+36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 36 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±9,±18,±36,±\frac{9}{2},±3,±6,±12,±\frac{9}{4},±\frac{3}{2},±1,±2,±4,±\frac{3}{4},±\frac{1}{2},±\frac{1}{4}$$

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 4x^{4}-40x^{3}-45x^{2}+725x+36 by x+4 to get 4x^{3}-56x^{2}+179x+9. Solve the equation where the result equals to 0.

$$4x^{3}-56x^{2}+179x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=9$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-56x^{2}+179x+9 by x-9 to get 4x^{2}-20x-1. Solve the equation where the result equals to 0.

$$4x^{2}-20x-1=0$$

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 4 for a, -20 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{20±4\sqrt{26}}{8}$$

Solve the equation 4x^{2}-20x-1=0 when ± is plus and when ± is minus.

$$x=\frac{5-\sqrt{26}}{2} x=\frac{\sqrt{26}+5}{2}$$

List all found solutions.

$$x=-4 x=9 x=\frac{5-\sqrt{26}}{2} x=\frac{\sqrt{26}+5}{2}$$

Substitute -4 for x in the equation 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\left(-4\right)^{2}-10\left(-4\right)=6+11\sqrt{\left(-4\right)^{2}-5\left(-4\right)}$$

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

$$72=72$$

Substitute 9 for x in the equation 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times 9^{2}-10\times 9=6+11\sqrt{9^{2}-5\times 9}$$

Simplify. The value x=9 satisfies the equation.

$$72=72$$

Substitute \frac{5-\sqrt{26}}{2} for x in the equation 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times \left(\frac{5-\sqrt{26}}{2}\right)^{2}-10\times \frac{5-\sqrt{26}}{2}=6+11\sqrt{\left(\frac{5-\sqrt{26}}{2}\right)^{2}-5\times \frac{5-\sqrt{26}}{2}}$$

Simplify. The value x=\frac{5-\sqrt{26}}{2} does not satisfy the equation.

$$\frac{1}{2}=\frac{23}{2}$$

Substitute \frac{\sqrt{26}+5}{2} for x in the equation 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times \left(\frac{\sqrt{26}+5}{2}\right)^{2}-10\times \frac{\sqrt{26}+5}{2}=6+11\sqrt{\left(\frac{\sqrt{26}+5}{2}\right)^{2}-5\times \frac{\sqrt{26}+5}{2}}$$

Simplify. The value x=\frac{\sqrt{26}+5}{2} does not satisfy the equation.

$$\frac{1}{2}=\frac{23}{2}$$

List all solutions of 2x^{2}-10x-6=11\sqrt{x^{2}-5x}.

$$x=-4 x=9$$