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simonmagusflies

Member

Registered: 2021-05-23

Posts: 32

A question that has both my parents stumped! Quadratics.

(ax + 3)(5x^2 - bx + 4) = 20x^3 - 9x^2 - 2x + 12
The equation above is true for all x, where a and b are constants. What is the value of ab?
A) 18
B) 20
C) 24
D) 40

The answer is C, 24. For the first term, you multiply (ax + 3) into the second bracket term to get 5ax^3 + 15x^2 - abx^2 - 3bx + 4ax + 12 = 9x^2 - 2x + 12. That much I got. But, the next step is to eliminate all numbers with non-x^2 terms?? So you get 15x^2 - abx&^2 = 9x^2. THAT I don't get. Neither do my parents. Can someone please explain what reasoning they used to eliminate all numbers with non-x^2 terms?

Thank you!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: A question that has both my parents stumped! Quadratics.

hi simonmagusflies

5ax^3 + 15x^2 - abx^2 - 3bx + 4ax + 12 = 9x^2 - 2x + 12.

You've 'lost' the x-cubed term. 

By comparing the coefficients of x^3, x^2 and x, it must be possible to find two unknowns.  But ..... why not

ax times 5x^2 makes 20x^3 so a= 4.

Now compare the x^2 terms:

-ab x^2 + 15x^2 = - 9x^2 .... so -ab = -9 - 15 ...... ab = 24.

Strictly I ought to check that the x coefficients work out too.  I'll leave that as an exercise.

Bob

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