Solve x^2+left(3.5-xright)^2=9

$${ x }^{ 2 } + { \left(3.5-x \right) }^{ 2 } =9$$

Solve for x

$$x = \frac{\sqrt{23} + 7}{4} \approx 2.948957881$$

$$x=\frac{7-\sqrt{23}}{4}\approx 0.551042119$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3.5-x\right)^{2}.

$$x^{2}+12.25-7x+x^{2}=9$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+12.25-7x=9$$

Subtract 9 from both sides.

$$2x^{2}+12.25-7x-9=0$$

Subtract 9 from 12.25 to get 3.25.

$$2x^{2}+3.25-7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-7x+3.25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 3.25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 3.25}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 3.25}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 3.25}}{2\times 2}$$

Multiply -8 times 3.25.

$$x=\frac{-\left(-7\right)±\sqrt{49-26}}{2\times 2}$$

Add 49 to -26.

$$x=\frac{-\left(-7\right)±\sqrt{23}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{23}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{23}}{4}$$

Now solve the equation x=\frac{7±\sqrt{23}}{4} when ± is plus. Add 7 to \sqrt{23}.

$$x=\frac{\sqrt{23}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{23}}{4} when ± is minus. Subtract \sqrt{23} from 7.

$$x=\frac{7-\sqrt{23}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{23}+7}{4} x=\frac{7-\sqrt{23}}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3.5-x\right)^{2}.

$$x^{2}+12.25-7x+x^{2}=9$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+12.25-7x=9$$

Subtract 12.25 from both sides.

$$2x^{2}-7x=9-12.25$$

Subtract 12.25 from 9 to get -3.25.

$$2x^{2}-7x=-3.25$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=-\frac{3.25}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=-\frac{3.25}{2}$$

Divide -3.25 by 2.

$$x^{2}-\frac{7}{2}x=-1.625$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-1.625+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-1.625+\frac{49}{16}$$

Add -1.625 to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{23}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{23}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{23}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{23}}{4} x-\frac{7}{4}=-\frac{\sqrt{23}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{23}+7}{4} x=\frac{7-\sqrt{23}}{4}$$