Solve a^2-9a+20

$$a ^ { 2 } - 9 a + 20$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+20. To find p and q, set up a system to be solved.

$$p+q=-9 pq=1\times 20=20$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q 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.

$$p=-5 q=-4$$

Rewrite a^{2}-9a+20 as \left(a^{2}-5a\right)+\left(-4a+20\right).

$$\left(a^{2}-5a\right)+\left(-4a+20\right)$$

Factor out a in the first and -4 in the second group.

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

Factor out common term a-5 by using distributive property.

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$a^{2}-9a+20=0$$

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.

$$a=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20}}{2}$$

Square -9.

$$a=\frac{-\left(-9\right)±\sqrt{81-4\times 20}}{2}$$

Multiply -4 times 20.

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

Add 81 to -80.

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

Take the square root of 1.

$$a=\frac{-\left(-9\right)±1}{2}$$

The opposite of -9 is 9.

$$a=\frac{9±1}{2}$$

Now solve the equation a=\frac{9±1}{2} when ± is plus. Add 9 to 1.

$$a=\frac{10}{2}$$

Divide 10 by 2.

$$a=5$$

Now solve the equation a=\frac{9±1}{2} when ± is minus. Subtract 1 from 9.

$$a=\frac{8}{2}$$

Divide 8 by 2.

$$a=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and 4 for x_{2}.

$$a^{2}-9a+20=\left(a-5\right)\left(a-4\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -9x +20 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 9 rs = 20$$

Two numbers r and s sum up to 9 exactly when the average of the two numbers is \frac{1}{2}*9 = \frac{9}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{9}{2} - u s = \frac{9}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 20

$$(\frac{9}{2} - u) (\frac{9}{2} + u) = 20$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{81}{4} - u^2 = 20$$

Simplify the expression by subtracting \frac{81}{4} on both sides

$$-u^2 = 20-\frac{81}{4} = -\frac{1}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{9}{2} - \frac{1}{2} = 4 s = \frac{9}{2} + \frac{1}{2} = 5$$

Evaluate

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