Solve 1/x+3+7-2x/x-5=7-x^2/x^2-2x-15

$$\frac{ 1 }{ x+3 } + \frac{ 7-2x }{ x-5 } = \frac{ 7- { x }^{ 2 } }{ { x }^{ 2 } -2x-15 }$$

Solve for x

$$x=\sqrt{10}+1\approx 4.16227766$$

$$x=1-\sqrt{10}\approx -2.16227766$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+3\right), the least common multiple of x+3,x-5,x^{2}-2x-15.

$$x-5+\left(x+3\right)\left(7-2x\right)=7-x^{2}$$

Use the distributive property to multiply x+3 by 7-2x and combine like terms.

$$x-5+x-2x^{2}+21=7-x^{2}$$

Combine x and x to get 2x.

$$2x-5-2x^{2}+21=7-x^{2}$$

Add -5 and 21 to get 16.

$$2x+16-2x^{2}=7-x^{2}$$

Subtract 7 from both sides.

$$2x+16-2x^{2}-7=-x^{2}$$

Subtract 7 from 16 to get 9.

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

Add x^{2} to both sides.

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

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

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-2±\sqrt{4+36}}{2\left(-1\right)}$$

Add 4 to 36.

$$x=\frac{-2±\sqrt{40}}{2\left(-1\right)}$$

Take the square root of 40.

$$x=\frac{-2±2\sqrt{10}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±2\sqrt{10}}{-2}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{-2} when ± is plus. Add -2 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{-2}$$

Divide -2+2\sqrt{10} by -2.

$$x=1-\sqrt{10}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{-2} when ± is minus. Subtract 2\sqrt{10} from -2.

$$x=\frac{-2\sqrt{10}-2}{-2}$$

Divide -2-2\sqrt{10} by -2.

$$x=\sqrt{10}+1$$

The equation is now solved.

$$x=1-\sqrt{10} x=\sqrt{10}+1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+3\right), the least common multiple of x+3,x-5,x^{2}-2x-15.

$$x-5+\left(x+3\right)\left(7-2x\right)=7-x^{2}$$

Use the distributive property to multiply x+3 by 7-2x and combine like terms.

$$x-5+x-2x^{2}+21=7-x^{2}$$

Combine x and x to get 2x.

$$2x-5-2x^{2}+21=7-x^{2}$$

Add -5 and 21 to get 16.

$$2x+16-2x^{2}=7-x^{2}$$

Add x^{2} to both sides.

$$2x+16-2x^{2}+x^{2}=7$$

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

$$2x+16-x^{2}=7$$

Subtract 16 from both sides.

$$2x-x^{2}=7-16$$

Subtract 16 from 7 to get -9.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}+2x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{2}{-1}x=-\frac{9}{-1}$$

Divide 2 by -1.

$$x^{2}-2x=-\frac{9}{-1}$$

Divide -9 by -1.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=9+1$$

Add 9 to 1.

$$x^{2}-2x+1=10$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-1=\sqrt{10} x-1=-\sqrt{10}$$

Add 1 to both sides of the equation.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$