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worty [1.4K]
3 years ago
5

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 $a b c$?
Mathematics
2 answers:
Natalka [10]3 years ago
7 0

Answer: a=6 b=9 c=8

Step-by-step explanation:

The problem consists of finding the roots of the quadratic equations:

x^2+15x+54=0\\

x^2-17x+72=0\\

The roots can be found with the following equation for solving quadratic equations:

x_{12}=\frac{-B\pm\sqrt{B^2-4AC}}{2A}  for equation: Ax^2+Bx+C=0

After solving the equations you can write the result as:

x^2+15x+54=(x+6)(x+9)\\

x^2-17x+72=(x-9)(x-8)

Naya [18.7K]3 years ago
7 0

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

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