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

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

Solve for x

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

$$x=1$$

Steps Using Factoring By Grouping

Combine 5x^{2} and -9x^{2} to get -4x^{2}.

$$-4x^{2}-5x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-4\times 9=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=4 b=-9$$

Rewrite -4x^{2}-5x+9 as \left(-4x^{2}+4x\right)+\left(-9x+9\right).

$$\left(-4x^{2}+4x\right)+\left(-9x+9\right)$$

Factor out 4x in the first and 9 in the second group.

$$4x\left(-x+1\right)+9\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(4x+9\right)$$

To find equation solutions, solve -x+1=0 and 4x+9=0.

$$x=1 x=-\frac{9}{4}$$

Steps Using the Quadratic Formula

Combine 5x^{2} and -9x^{2} to get -4x^{2}.

$$-4x^{2}-5x+9=0$$

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+16\times 9}}{2\left(-4\right)}$$

Multiply 16 times 9.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\left(-4\right)}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\left(-4\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{5±13}{-8}$$

Now solve the equation x=\frac{5±13}{-8} when ± is plus. Add 5 to 13.

$$x=\frac{18}{-8}$$

Reduce the fraction \frac{18}{-8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{4}$$

Now solve the equation x=\frac{5±13}{-8} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

The equation is now solved.

$$x=-\frac{9}{4} x=1$$

Steps for Completing the Square

Combine 5x^{2} and -9x^{2} to get -4x^{2}.

$$-4x^{2}-5x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}-5x=-9$$

Divide both sides by -4.

$$\frac{-4x^{2}-5x}{-4}=-\frac{9}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{5}{-4}\right)x=-\frac{9}{-4}$$

Divide -5 by -4.

$$x^{2}+\frac{5}{4}x=-\frac{9}{-4}$$

Divide -9 by -4.

$$x^{2}+\frac{5}{4}x=\frac{9}{4}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{9}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{9}{4}+\frac{25}{64}$$

Add \frac{9}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{169}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{13}{8} x+\frac{5}{8}=-\frac{13}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=1 x=-\frac{9}{4}$$