Solve 2x-15/x+7=0

$$2 x - \frac { 15 } { x } + 7 = 0$$

Solve for x

$$x=-5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2xx-15+x\times 7=0$$

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

$$2x^{2}-15+x\times 7=0$$

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

$$2x^{2}+7x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\left(-15\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 2x^{2}+7x-15 as \left(2x^{2}-3x\right)+\left(10x-15\right).

$$\left(2x^{2}-3x\right)+\left(10x-15\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(2x-3\right)+5\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+5\right)$$

To find equation solutions, solve 2x-3=0 and x+5=0.

$$x=\frac{3}{2} x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2xx-15+x\times 7=0$$

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

$$2x^{2}-15+x\times 7=0$$

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.

$$2x^{2}+7x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-7±\sqrt{49+120}}{2\times 2}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±13}{4}$$

Now solve the equation x=\frac{-7±13}{4} when ± is plus. Add -7 to 13.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{-7±13}{4} when ± is minus. Subtract 13 from -7.

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

Divide -20 by 4.

$$x=-5$$

The equation is now solved.

$$x=\frac{3}{2} x=-5$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2xx-15+x\times 7=0$$

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

$$2x^{2}-15+x\times 7=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$2x^{2}+x\times 7=15$$

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.

$$2x^{2}+7x=15$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=\frac{15}{2}$$

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.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=\frac{15}{2}+\left(\frac{7}{4}\right)^{2}$$

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

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$$

Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}$$

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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{13}{4} x+\frac{7}{4}=-\frac{13}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{3}{2} x=-5$$