Solve 4m^2+5=9m

$$4 m ^ { 2 } + 5 = 9 m$$

Solve for m

$$m=1$$

$$m = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

Subtract 9m from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4m^{2}+am+bm+5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=4\times 5=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

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

$$\left(4m^{2}-5m\right)+\left(-4m+5\right)$$

Factor out m in the first and -1 in the second group.

$$m\left(4m-5\right)-\left(4m-5\right)$$

Factor out common term 4m-5 by using distributive property.

$$\left(4m-5\right)\left(m-1\right)$$

To find equation solutions, solve 4m-5=0 and m-1=0.

$$m=\frac{5}{4} m=1$$

Steps Using the Quadratic Formula

Subtract 9m from both sides.

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

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.

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

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

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

Square -9.

$$m=\frac{-\left(-9\right)±\sqrt{81-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$m=\frac{-\left(-9\right)±\sqrt{81-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$m=\frac{-\left(-9\right)±\sqrt{81-80}}{2\times 4}$$

Add 81 to -80.

$$m=\frac{-\left(-9\right)±\sqrt{1}}{2\times 4}$$

Take the square root of 1.

$$m=\frac{-\left(-9\right)±1}{2\times 4}$$

The opposite of -9 is 9.

$$m=\frac{9±1}{2\times 4}$$

Multiply 2 times 4.

$$m=\frac{9±1}{8}$$

Now solve the equation m=\frac{9±1}{8} when ± is plus. Add 9 to 1.

$$m=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$m=\frac{5}{4}$$

Now solve the equation m=\frac{9±1}{8} when ± is minus. Subtract 1 from 9.

$$m=\frac{8}{8}$$

Divide 8 by 8.

$$m=1$$

The equation is now solved.

$$m=\frac{5}{4} m=1$$

Steps for Completing the Square

Subtract 9m from both sides.

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

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

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{9}{4}m+\frac{81}{64}=-\frac{5}{4}+\frac{81}{64}$$

Add -\frac{5}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}-\frac{9}{4}m+\frac{81}{64}=\frac{1}{64}$$

Factor m^{2}-\frac{9}{4}m+\frac{81}{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}.

$$\left(m-\frac{9}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{9}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$m-\frac{9}{8}=\frac{1}{8} m-\frac{9}{8}=-\frac{1}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$m=\frac{5}{4} m=1$$