Big Book Test-6, Section 4

doubt_1_algebra

I come across this type of question pretty often, and while my way of solving is correct and gives me the right answer, it’s a bit tedious and I was wondering if there is a easier method

My method:
if x=3 is one solution, then the equation must be divisible by (x-3)
The I basically do long division of x^2 + rx -24 by x -3
The remainder I get is 3r - 15
As x-3 perfectly divides the equation 3r - 15 =0
x = 5

You can just substitute x=3 in the equation right?

9 + 3r - 20 = 4 → 3r = 15 → r=5

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Why didn’t I think of this? * Facepalm*

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