Solve left(x-3right)^2=11x-3

$${ \left(x-3 \right) }^{ 2 } =11x-3$$

Solve for x

$$x = \frac{\sqrt{241} + 17}{2} \approx 16.262087348$$

$$x=\frac{17-\sqrt{241}}{2}\approx 0.737912652$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=11x-3$$

Subtract 11x from both sides.

$$x^{2}-6x+9-11x=-3$$

Combine -6x and -11x to get -17x.

$$x^{2}-17x+9=-3$$

Add 3 to both sides.

$$x^{2}-17x+9+3=0$$

Add 9 and 3 to get 12.

$$x^{2}-17x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -17 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 12}}{2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-17\right)±\sqrt{289-48}}{2}$$

Add 289 to -48.

$$x=\frac{-\left(-17\right)±\sqrt{241}}{2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{241}}{2}$$

Now solve the equation x=\frac{17±\sqrt{241}}{2} when ± is plus. Add 17 to \sqrt{241}.

$$x=\frac{\sqrt{241}+17}{2}$$

Now solve the equation x=\frac{17±\sqrt{241}}{2} when ± is minus. Subtract \sqrt{241} from 17.

$$x=\frac{17-\sqrt{241}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{241}+17}{2} x=\frac{17-\sqrt{241}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=11x-3$$

Subtract 11x from both sides.

$$x^{2}-6x+9-11x=-3$$

Combine -6x and -11x to get -17x.

$$x^{2}-17x+9=-3$$

Subtract 9 from both sides.

$$x^{2}-17x=-3-9$$

Subtract 9 from -3 to get -12.

$$x^{2}-17x=-12$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=-12+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=-12+\frac{289}{4}$$

Add -12 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{241}{4}$$

Factor x^{2}-17x+\frac{289}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{241}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{241}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{\sqrt{241}}{2} x-\frac{17}{2}=-\frac{\sqrt{241}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{\sqrt{241}+17}{2} x=\frac{17-\sqrt{241}}{2}$$