Solve 9/2x^2+17x=1

$$\frac { 9 } { 2 x ^ { 2 } + 17 x } = 1$$

Solve for x

$$x=-9$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -\frac{17}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+17\right).

$$9=x\left(2x+17\right)$$

Use the distributive property to multiply x by 2x+17.

$$9=2x^{2}+17x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+17x=9$$

Subtract 9 from both sides.

$$2x^{2}+17x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=17 ab=2\left(-9\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 17.

$$a=-1 b=18$$

Rewrite 2x^{2}+17x-9 as \left(2x^{2}-x\right)+\left(18x-9\right).

$$\left(2x^{2}-x\right)+\left(18x-9\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(2x-1\right)+9\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+9\right)$$

To find equation solutions, solve 2x-1=0 and x+9=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{17}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+17\right).

$$9=x\left(2x+17\right)$$

Use the distributive property to multiply x by 2x+17.

$$9=2x^{2}+17x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+17x=9$$

Subtract 9 from both sides.

$$2x^{2}+17x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 17 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-17±\sqrt{289-8\left(-9\right)}}{2\times 2}$$

Multiply -8 times -9.

$$x=\frac{-17±\sqrt{289+72}}{2\times 2}$$

Add 289 to 72.

$$x=\frac{-17±\sqrt{361}}{2\times 2}$$

Take the square root of 361.

$$x=\frac{-17±19}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-17±19}{4}$$

Now solve the equation x=\frac{-17±19}{4} when ± is plus. Add -17 to 19.

$$x=\frac{2}{4}$$

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

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

Now solve the equation x=\frac{-17±19}{4} when ± is minus. Subtract 19 from -17.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{17}{2},0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x+17\right).

$$9=x\left(2x+17\right)$$

Use the distributive property to multiply x by 2x+17.

$$9=2x^{2}+17x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+17x=9$$

Divide both sides by 2.

$$\frac{2x^{2}+17x}{2}=\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{17}{2}x=\frac{9}{2}$$

Divide \frac{17}{2}, the coefficient of the x term, by 2 to get \frac{17}{4}. Then add the square of \frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{2}x+\left(\frac{17}{4}\right)^{2}=\frac{9}{2}+\left(\frac{17}{4}\right)^{2}$$

Square \frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{9}{2}+\frac{289}{16}$$

Add \frac{9}{2} to \frac{289}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{361}{16}$$

Factor x^{2}+\frac{17}{2}x+\frac{289}{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}.

$$\left(x+\frac{17}{4}\right)^{2}=\frac{361}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{4}\right)^{2}}=\sqrt{\frac{361}{16}}$$

Simplify.

$$x+\frac{17}{4}=\frac{19}{4} x+\frac{17}{4}=-\frac{19}{4}$$

Subtract \frac{17}{4} from both sides of the equation.

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