Solve left(4+xright)^2+left(3+xright)^2=49

$${ \left(4+x \right) }^{ 2 } + { \left(3+x \right) }^{ 2 } = 49$$

Solve for x

$$x = \frac{\sqrt{97} - 7}{2} \approx 1.424428901$$

$$x=\frac{-\sqrt{97}-7}{2}\approx -8.424428901$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+x\right)^{2}.

$$16+8x+x^{2}+\left(3+x\right)^{2}=49$$

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

$$16+8x+x^{2}+9+6x+x^{2}=49$$

Add 16 and 9 to get 25.

$$25+8x+x^{2}+6x+x^{2}=49$$

Combine 8x and 6x to get 14x.

$$25+14x+x^{2}+x^{2}=49$$

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

$$25+14x+2x^{2}=49$$

Subtract 49 from both sides.

$$25+14x+2x^{2}-49=0$$

Subtract 49 from 25 to get -24.

$$-24+14x+2x^{2}=0$$

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.

$$2x^{2}+14x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 14 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-14±\sqrt{196-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$x=\frac{-14±\sqrt{196+192}}{2\times 2}$$

Add 196 to 192.

$$x=\frac{-14±\sqrt{388}}{2\times 2}$$

Take the square root of 388.

$$x=\frac{-14±2\sqrt{97}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-14±2\sqrt{97}}{4}$$

Now solve the equation x=\frac{-14±2\sqrt{97}}{4} when ± is plus. Add -14 to 2\sqrt{97}.

$$x=\frac{2\sqrt{97}-14}{4}$$

Divide -14+2\sqrt{97} by 4.

$$x=\frac{\sqrt{97}-7}{2}$$

Now solve the equation x=\frac{-14±2\sqrt{97}}{4} when ± is minus. Subtract 2\sqrt{97} from -14.

$$x=\frac{-2\sqrt{97}-14}{4}$$

Divide -14-2\sqrt{97} by 4.

$$x=\frac{-\sqrt{97}-7}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{97}-7}{2} x=\frac{-\sqrt{97}-7}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+x\right)^{2}.

$$16+8x+x^{2}+\left(3+x\right)^{2}=49$$

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

$$16+8x+x^{2}+9+6x+x^{2}=49$$

Add 16 and 9 to get 25.

$$25+8x+x^{2}+6x+x^{2}=49$$

Combine 8x and 6x to get 14x.

$$25+14x+x^{2}+x^{2}=49$$

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

$$25+14x+2x^{2}=49$$

Subtract 25 from both sides.

$$14x+2x^{2}=49-25$$

Subtract 25 from 49 to get 24.

$$14x+2x^{2}=24$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+14x=24$$

Divide both sides by 2.

$$\frac{2x^{2}+14x}{2}=\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{14}{2}x=\frac{24}{2}$$

Divide 14 by 2.

$$x^{2}+7x=\frac{24}{2}$$

Divide 24 by 2.

$$x^{2}+7x=12$$

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

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=12+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=12+\frac{49}{4}$$

Add 12 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{97}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{97}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{97}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{97}}{2} x+\frac{7}{2}=-\frac{\sqrt{97}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=\frac{\sqrt{97}-7}{2} x=\frac{-\sqrt{97}-7}{2}$$