SAT Math 5.5 Algebra and Functions
|Algebra and Functions||Solving Quadratic Equations by Factoring|
|Functions||Build new functions from existing functions|
|Passport to Advanced Math||Solving quadratic equations|
|Product Type||SAT Math|
|SAT Math||Algebra and Functions|
Well, the first function we’re given, x squared minus 100 over x squared plus 10x, can be factored.
X squared minus 100 looks a heck of a lot like the difference of two squares….
so the numerator becomes (x + 10) times (x - 10). And the denominator is x squared plus 10x…
which becomes x times (x + 10). There’s an x + 10 now in both the numerator
and denominator, so we can cancel out the x +10 in each place.
Our new equation looks like this: f of x equals x minus 10 over x.
We’re given that f of x minus 5 equals 1/3, so substituting x minus 5 into the first function,
we get f of x minus 5… equals x minus 5 minus 10…all over x minus 5… equals 1/3.
X minus 5 minus 10 simplifies to x minus 15. So x minus 15 over x minus 5 equals 1 over 3.
To eliminate the fractions on both sides,
we need to multiply both sides by (x - 5) and 3… to arrive at 3 times (x - 15) = x - 5.
We distribute the 3 so that 3x - 45 = x – 5…and solve.
And x equals 20. Our answer’s E.
And our messy functions are no more. Thank you, Mr. Clean.