Here’s How to Crack It
We are given a quadratic equation that contains two variables.
In this case, work with 2x − 3y = 5. If you square the left side of the equations, you get
(2x − 3y)^2 = 4x^2 – 12xy + 9y^2That’s precisely the expression for which you need to find the value. It’s also the third of the equations
from the box. Now, since you squared the left side, all you need to do is square the 5 on the right side of
the equation to discover that the expression equals 25, (C).
Did you notice that this question was just another version of being asked to solve for the value of an
expression rather than for a variable? Quadratics are one of ETS’s favorite ways to do that.
Solving Quadratics Set to Zero
Before factoring most quadratics, you need to set the equation equal to zero. Why? Well, if ab = 0, what
do you know about a and b? At least one of them must equal 0, right? That’s the key fact you need to solve
most quadratics.
Here’s an example:
9.If 3 – x + 7, and x ≠ 0, which of the following is a possible value for x ?A) −7
B) −1
C) 1
D) 3Here’s How to Crack It
The test writers have tried to hide that the equation is actually a quadratic. Start by multiplying both sides
of the equation by x to get rid of the fraction.
x = x(x + 7)3 x − 3 = x^2 + 7xNow, just rearrange the terms to set the quadratic equal to 0. You’ll get x^2 + 4x + 3 = 0. Now, it’s time to
factor:
x^2 + 4x + 3 = (x + 1)(x + 3) = 0