Solve x^2-5/x-1=7x+10/9

$$\frac { x ^ { 2 } - 5 } { x - 1 } = \frac { 7 x + 10 } { 9 }$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right), the least common multiple of x-1,9.

$$9\left(x^{2}-5\right)=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply 9 by x^{2}-5.

$$9x^{2}-45=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply x-1 by 7x+10 and combine like terms.

$$9x^{2}-45=7x^{2}+3x-10$$

Subtract 7x^{2} from both sides.

$$9x^{2}-45-7x^{2}=3x-10$$

Combine 9x^{2} and -7x^{2} to get 2x^{2}.

$$2x^{2}-45=3x-10$$

Subtract 3x from both sides.

$$2x^{2}-45-3x=-10$$

Add 10 to both sides.

$$2x^{2}-45-3x+10=0$$

Add -45 and 10 to get -35.

$$2x^{2}-35-3x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}-3x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-35\right)=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -70.

$$1,-70 2,-35 5,-14 7,-10$$

Calculate the sum for each pair.

$$1-70=-69 2-35=-33 5-14=-9 7-10=-3$$

The solution is the pair that gives sum -3.

$$a=-10 b=7$$

Rewrite 2x^{2}-3x-35 as \left(2x^{2}-10x\right)+\left(7x-35\right).

$$\left(2x^{2}-10x\right)+\left(7x-35\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-5\right)+7\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(2x+7\right)$$

To find equation solutions, solve x-5=0 and 2x+7=0.

$$x=5 x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right), the least common multiple of x-1,9.

$$9\left(x^{2}-5\right)=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply 9 by x^{2}-5.

$$9x^{2}-45=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply x-1 by 7x+10 and combine like terms.

$$9x^{2}-45=7x^{2}+3x-10$$

Subtract 7x^{2} from both sides.

$$9x^{2}-45-7x^{2}=3x-10$$

Combine 9x^{2} and -7x^{2} to get 2x^{2}.

$$2x^{2}-45=3x-10$$

Subtract 3x from both sides.

$$2x^{2}-45-3x=-10$$

Add 10 to both sides.

$$2x^{2}-45-3x+10=0$$

Add -45 and 10 to get -35.

$$2x^{2}-35-3x=0$$

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.

$$2x^{2}-3x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-35\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-35\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-35\right)}}{2\times 2}$$

Multiply -8 times -35.

$$x=\frac{-\left(-3\right)±\sqrt{9+280}}{2\times 2}$$

Add 9 to 280.

$$x=\frac{-\left(-3\right)±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$x=\frac{-\left(-3\right)±17}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±17}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±17}{4}$$

Now solve the equation x=\frac{3±17}{4} when ± is plus. Add 3 to 17.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{3±17}{4} when ± is minus. Subtract 17 from 3.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=5 x=-\frac{7}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right), the least common multiple of x-1,9.

$$9\left(x^{2}-5\right)=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply 9 by x^{2}-5.

$$9x^{2}-45=\left(x-1\right)\left(7x+10\right)$$

Use the distributive property to multiply x-1 by 7x+10 and combine like terms.

$$9x^{2}-45=7x^{2}+3x-10$$

Subtract 7x^{2} from both sides.

$$9x^{2}-45-7x^{2}=3x-10$$

Combine 9x^{2} and -7x^{2} to get 2x^{2}.

$$2x^{2}-45=3x-10$$

Subtract 3x from both sides.

$$2x^{2}-45-3x=-10$$

Add 45 to both sides.

$$2x^{2}-3x=-10+45$$

Add -10 and 45 to get 35.

$$2x^{2}-3x=35$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=\frac{35}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=\frac{35}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{35}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{35}{2}+\frac{9}{16}$$

Add \frac{35}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{289}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{289}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{289}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{17}{4} x-\frac{3}{4}=-\frac{17}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=5 x=-\frac{7}{2}$$