Question

Use two different methods to find the roots of the quadratic equation
h (x) = x2 + 3x - 18. Complete the explanation of the methods you used, and
why. Did you get the same roots?
I used the factoring method to first solve the equation, because I noticed that 6
and -3 have a sum of 3, and a product of -18. That gave the factorization of
(x + 6) and (x – 3). When setting both sets of parentheses equal to zero, the final
answers were r = and x =
=
=
The quadratic equation can also be used: a = 1, b = 3, and c = -18.
2
-b yb - 4 ac -3+ V3 - 4(1)(-18) -3 +9
XE
2a
2(1)
2
6
-12
which gives
and
2
2
:
The answers are (select)

Answer 1

Solution

Step 1

Write the equation

$h(x)\text{ = x}^2\text{ + 3x - 18}$

Step 2

Use the factorization method

$\begin{gathered} Equate\text{ the function to zero to find the roots.} \\ x^2\text{ + 3x - 18 = 0} \\ x^2\text{ + 6x - 3x - 18 = 0} \\ x(x\text{ + 6\rparen -3\lparen x + 6\rparen = 0} \\ (x\text{ + 6\rparen\lparen x - 3\rparen = 0} \\ \text{x + 6 = 0, x - 3 = 0} \\ x\text{ = -6, x = 3} \end{gathered}$

Step 3

Use the quadratic formula to find the roots of the function

$\begin{gathered} \text{From y = ax}^2\text{ + bx + c} \\ \text{a =1, b = 3 , c = -18} \\ \text{x = }\frac{-b\pm\sqrt{b^2\text{ - 4ac}}}{2a} \\ \text{x = }\frac{-3\pm\sqrt{3^2\text{ - 4}\times1\times(-18)}}{2\times1} \\ \text{x = }\frac{-3\text{ }\pm\sqrt{9\text{ + 72}}}{2} \\ x\text{ = }\frac{-3\text{ }\pm\text{ }\sqrt{81}}{2} \\ \text{x = }\frac{-3\text{ }\pm\text{ 9}}{2} \\ \text{x = }\frac{-3-9}{2}\text{ , x = }\frac{-3+9}{2} \\ \text{x = -6 , x = 3} \end{gathered}$