Solve 9x^2-14x^+5=0

$$9 { x }^{ 2 } -14 { x }^{ } +5 = 0$$

Solve for x

$$x=1$$

$$x=\frac{5}{9}\approx 0.555555556$$

Steps Using Factoring By Grouping

Calculate x to the power of 1 and get x.

$$9x^{2}-14x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-14 ab=9\times 5=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite 9x^{2}-14x+5 as \left(9x^{2}-9x\right)+\left(-5x+5\right).

$$\left(9x^{2}-9x\right)+\left(-5x+5\right)$$

Factor out 9x in the first and -5 in the second group.

$$9x\left(x-1\right)-5\left(x-1\right)$$

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

$$\left(x-1\right)\left(9x-5\right)$$

To find equation solutions, solve x-1=0 and 9x-5=0.

$$x=1 x=\frac{5}{9}$$

Steps Using the Quadratic Formula

Calculate x to the power of 1 and get x.

$$9x^{2}-14x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -14 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 9\times 5}}{2\times 9}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 9\times 5}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-14\right)±\sqrt{196-36\times 5}}{2\times 9}$$

Multiply -36 times 5.

$$x=\frac{-\left(-14\right)±\sqrt{196-180}}{2\times 9}$$

Add 196 to -180.

$$x=\frac{-\left(-14\right)±\sqrt{16}}{2\times 9}$$

Take the square root of 16.

$$x=\frac{-\left(-14\right)±4}{2\times 9}$$

The opposite of -14 is 14.

$$x=\frac{14±4}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{14±4}{18}$$

Now solve the equation x=\frac{14±4}{18} when ± is plus. Add 14 to 4.

$$x=\frac{18}{18}$$

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{14±4}{18} when ± is minus. Subtract 4 from 14.

$$x=\frac{10}{18}$$

Reduce the fraction \frac{10}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{9}$$

The equation is now solved.

$$x=1 x=\frac{5}{9}$$

Steps for Completing the Square

Calculate x to the power of 1 and get x.

$$9x^{2}-14x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-14x=-5$$

Divide both sides by 9.

$$\frac{9x^{2}-14x}{9}=-\frac{5}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{14}{9}x=-\frac{5}{9}$$

Divide -\frac{14}{9}, the coefficient of the x term, by 2 to get -\frac{7}{9}. Then add the square of -\frac{7}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{9}x+\left(-\frac{7}{9}\right)^{2}=-\frac{5}{9}+\left(-\frac{7}{9}\right)^{2}$$

Square -\frac{7}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{9}x+\frac{49}{81}=-\frac{5}{9}+\frac{49}{81}$$

Add -\frac{5}{9} to \frac{49}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{9}x+\frac{49}{81}=\frac{4}{81}$$

Factor x^{2}-\frac{14}{9}x+\frac{49}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{9}\right)^{2}=\frac{4}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{9}\right)^{2}}=\sqrt{\frac{4}{81}}$$

Simplify.

$$x-\frac{7}{9}=\frac{2}{9} x-\frac{7}{9}=-\frac{2}{9}$$

Add \frac{7}{9} to both sides of the equation.

$$x=1 x=\frac{5}{9}$$