Answer: a=6 b=9 c=8
Step-by-step explanation:
The problem consists of finding the roots of the quadratic equations:
[tex]x^2+15x+54=0\\[/tex]
[tex]x^2-17x+72=0\\[/tex]
The roots can be found with the following equation for solving quadratic equations:
[tex]x_{12}=\frac{-B\pm\sqrt{B^2-4AC}}{2A}[/tex] for equation: [tex]Ax^2+Bx+C=0[/tex]
After solving the equations you can write the result as:
[tex]x^2+15x+54=(x+6)(x+9)\\[/tex]
[tex]x^2-17x+72=(x-9)(x-8)[/tex]
Answer:
b = 9
a = 6
c= 8
Therefore, a,b,c = 6,9,8
Question:
The expression $x^2 - 15x - 54$ can be written as $(x-a)(x-b),$ and the expression $x^2 - 17x + 72$ written as $(x - b)(x - c)$, where $a$, $b$, and $c$ are integers. What is the value of $a b c$?
Step-by-step explanation:
To determine the values of a,b and c.
We need factorize the two equations.
i) x^2 -15x +54
x^2 -6x -9x +54
x(x-6) -9(x-6)
=(x-6)(x-9)
ii) x^2 -17x +72
x^2 -9x -8x +72
x(x-9) -8(x-9)
=(x-9)(x-8)
From the question:
Comparing
(x-a)(x-b) to (x-6)(x-9)
And
(x-b)(x-c) to (x-9)(x-8)
We can see that b is common in the two cases and also 9 is common in the two factorised equations.
So,
b = 9
a = 6
c= 8
Therefore, a,b,c = 6,9,8
