Solve x+3/2=5/x+4

$$\frac { x + 3 } { 2 } = \frac { 5 } { x + 4 }$$

Solve for x

$$x=\frac{\sqrt{41}-7}{2}\approx -0.298437881$$

$$x=\frac{-\sqrt{41}-7}{2}\approx -6.701562119$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by 2\left(x+4\right), the least common multiple of 2,x+4.

$$\left(x+4\right)\left(x+3\right)=2\times 5$$

Use the distributive property to multiply x+4 by x+3 and combine like terms.

$$x^{2}+7x+12=2\times 5$$

Multiply 2 and 5 to get 10.

$$x^{2}+7x+12=10$$

Subtract 10 from both sides.

$$x^{2}+7x+12-10=0$$

Subtract 10 from 12 to get 2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8}}{2}$$

Add 49 to -8.

$$x=\frac{-7±\sqrt{41}}{2}$$

Now solve the equation x=\frac{-7±\sqrt{41}}{2} when ± is plus. Add -7 to \sqrt{41}.

$$x=\frac{\sqrt{41}-7}{2}$$

Now solve the equation x=\frac{-7±\sqrt{41}}{2} when ± is minus. Subtract \sqrt{41} from -7.

$$x=\frac{-\sqrt{41}-7}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-7}{2} x=\frac{-\sqrt{41}-7}{2}$$

Steps for Completing the Square

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by 2\left(x+4\right), the least common multiple of 2,x+4.

$$\left(x+4\right)\left(x+3\right)=2\times 5$$

Use the distributive property to multiply x+4 by x+3 and combine like terms.

$$x^{2}+7x+12=2\times 5$$

Multiply 2 and 5 to get 10.

$$x^{2}+7x+12=10$$

Subtract 12 from both sides.

$$x^{2}+7x=10-12$$

Subtract 12 from 10 to get -2.

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

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.

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

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

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

Add -2 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{41}{4}$$

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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{41}}{2} x+\frac{7}{2}=-\frac{\sqrt{41}}{2}$$

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

$$x=\frac{\sqrt{41}-7}{2} x=\frac{-\sqrt{41}-7}{2}$$