Answer:
[tex]$ 3x^2 + 27x + 54 = 0 $[/tex]
Step-by-step explanation:
A quadratic equation is given by: [tex]$ a_1 x^2 + ax + b = 0 $[/tex]
The leading co-efficient is [tex]$ a_1 $[/tex].
Given [tex]$ a_1 = 3$[/tex]. Therefore the quadratic equation becomes:
[tex]$ 3x^2 + ax + b = 0 $[/tex]. We are to find [tex]$ (a,b) $[/tex].
Given the solutions of this equation are: [tex]$ -3 $[/tex] and [tex]$ -6 $[/tex].
Substituting this in the equation we get:
[tex]$ 3(-3)^2 +a(-3) + b = 27 -3a + b = 0 \hspace{15mm} (1) $[/tex]
[tex]$ 3(-6)^2 +a(-6) + b = 108 -6a + b = 0 \hspace{15mm} (2) $[/tex]
Solving [tex]$ (1) $[/tex] and [tex]$ (2) $[/tex] we get
[tex]$ b = 54 $[/tex]. Substituting in [tex]$ (1) $[/tex] we get
[tex]$ a = 27 $[/tex]. Thus the equation becomes:
[tex]$ 3x^2 + 27x + 54 = 0 $[/tex]
