Solve 17x^2-32x+15=0

$$17 { x }^{ 2 } -32x+15=0$$

Solve for x

$$x=1$$

$$x=\frac{15}{17}\approx 0.882352941$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 17x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-32 ab=17\times 15=255$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 255.

$$-1,-255 -3,-85 -5,-51 -15,-17$$

Calculate the sum for each pair.

$$-1-255=-256 -3-85=-88 -5-51=-56 -15-17=-32$$

The solution is the pair that gives sum -32.

$$a=-17 b=-15$$

Rewrite 17x^{2}-32x+15 as \left(17x^{2}-17x\right)+\left(-15x+15\right).

$$\left(17x^{2}-17x\right)+\left(-15x+15\right)$$

Factor out 17x in the first and -15 in the second group.

$$17x\left(x-1\right)-15\left(x-1\right)$$

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

$$\left(x-1\right)\left(17x-15\right)$$

To find equation solutions, solve x-1=0 and 17x-15=0.

$$x=1 x=\frac{15}{17}$$

Steps Using the Quadratic Formula

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.

$$17x^{2}-32x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, -32 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 17\times 15}}{2\times 17}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 17\times 15}}{2\times 17}$$

Multiply -4 times 17.

$$x=\frac{-\left(-32\right)±\sqrt{1024-68\times 15}}{2\times 17}$$

Multiply -68 times 15.

$$x=\frac{-\left(-32\right)±\sqrt{1024-1020}}{2\times 17}$$

Add 1024 to -1020.

$$x=\frac{-\left(-32\right)±\sqrt{4}}{2\times 17}$$

Take the square root of 4.

$$x=\frac{-\left(-32\right)±2}{2\times 17}$$

The opposite of -32 is 32.

$$x=\frac{32±2}{2\times 17}$$

Multiply 2 times 17.

$$x=\frac{32±2}{34}$$

Now solve the equation x=\frac{32±2}{34} when ± is plus. Add 32 to 2.

$$x=\frac{34}{34}$$

Divide 34 by 34.

$$x=1$$

Now solve the equation x=\frac{32±2}{34} when ± is minus. Subtract 2 from 32.

$$x=\frac{30}{34}$$

Reduce the fraction \frac{30}{34} to lowest terms by extracting and canceling out 2.

$$x=\frac{15}{17}$$

The equation is now solved.

$$x=1 x=\frac{15}{17}$$

Steps for Completing the Square

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.

$$17x^{2}-32x+15=0$$

Subtract 15 from both sides of the equation.

$$17x^{2}-32x+15-15=-15$$

Subtracting 15 from itself leaves 0.

$$17x^{2}-32x=-15$$

Divide both sides by 17.

$$\frac{17x^{2}-32x}{17}=-\frac{15}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$x^{2}-\frac{32}{17}x=-\frac{15}{17}$$

Divide -\frac{32}{17}, the coefficient of the x term, by 2 to get -\frac{16}{17}. Then add the square of -\frac{16}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{17}x+\left(-\frac{16}{17}\right)^{2}=-\frac{15}{17}+\left(-\frac{16}{17}\right)^{2}$$

Square -\frac{16}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{17}x+\frac{256}{289}=-\frac{15}{17}+\frac{256}{289}$$

Add -\frac{15}{17} to \frac{256}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{17}x+\frac{256}{289}=\frac{1}{289}$$

Factor x^{2}-\frac{32}{17}x+\frac{256}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{17}\right)^{2}=\frac{1}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{17}\right)^{2}}=\sqrt{\frac{1}{289}}$$

Simplify.

$$x-\frac{16}{17}=\frac{1}{17} x-\frac{16}{17}=-\frac{1}{17}$$

Add \frac{16}{17} to both sides of the equation.

$$x=1 x=\frac{15}{17}$$