So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 18 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of Coordinate Geometry can be kind of tricky, but never fear, PrepScholar has got your back!

Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.

The questions asks us to find which points will intersect two separate graphs, and we’re provided the equations for these two graphs. So this sounds as if we’ll be tested on what we know about Coordinate Geometry. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.

What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

 

1. We have two equations in the $xy$-plane: $y=x+2$ and $y=x^2$
2. We want to know at what points these two graphs intersect

 

Develop a Plan

Wherever two graphs intersect, their $x$- and $y$-values must be equal. We already have equations solved for $y$. So let’s set them equal to each other, solve for $x$ to get the $x$-coordinates of intersection, then plug that $x$-value into one of the equations to get the $y$-coordinate of where the graphs intersect.

Solve the Question

Let’s set the two equations equal to each other.

$$x^2=x+2$$

Looks like we have a quadratic equation here. Let’s try to factor it. First, let’s rearrange it so that all terms are on the left side. Subtracting $x$ and $2$ from both sides we get:

$$x^2-x-2=0$$

Before we go any further, let’s briefly review how to factor quadratic equations.

 

Concept Refresher – Factoring a Quadratic Equation

On the GRE, we can expect to find quadratic equations that look like:

$$x^2+bx+c=0$$

Here, $b$ and $c$ are both integers. The most efficient way to solve these is to factor the quadratic equation. Our goal is to put it in the form: $(x+y)(x+z)=0$, where $y$ and $z$ are two integers. Once the equation is in that form, we know that if two numbers multiplied together give $0$ as a product, then one of the numbers must be $0$. So therefore either $x+y=0$, in which case $x=-y$, or $x+z=0$, in which case $x=-z$.

The easiest way to understand the rationale for how to find $y$ and $z$ is to actually work backwards, expanding $(x+y)(x+z)=0$ so that it looks like $x^2+ax+b=0$. To expand $(x+y)(x+z)=0$, we will use the FOIL method. The FOIL method tells use that for the four terms $x, y, x, \and z$, we need to multiply together the First term within each set of parentheses, then the Outer terms, then the Inner terms, and finally the Last terms. Doing so gives us:

$(x+y)(x+z)$	$=$	$x·x+x·z+y·x+y·z$
$ $	$ $	$ $
$(x+y)(x+z)$	$=$	$x^2+xz+yx+yz$

We can then cleverly combine the two middle terms, $xz+yx$ and write them as $(y+z)x$. Doing so allows us to see how this factored form is related to the original quadratic equation. Let’s write them next to each other to look at how they’re related.

$$x^2+(b)x+c=x^2+(y+z)x+yz$$

Ah! So here we can see that the $b$ seems to match up with the $(y+z)$ and the $c$ matches up with $yz$. After all, for the equation to hold, both sides of the equation must be equal, so anything multiplying the $x$ on the left side of the equation must be equal to whatever is multiplying the $x$ on the right side of the equation. Then the $x^2$ is the same on both sides, so $c$ must be equal to $yz$.

So from this we can see that $b=y+z$ and also that $c=yz$. So if we think about going from the original quadratic equation that gave us $b$ and $c$ and writing it in the form $(x+y)(x+z)=0$, we see that $b$ must be the sum of $y$ and $z$ and $c$ must be the product of $y$ and $z$.

Let’s try an example. Say we want to factor $x^2+5x+6=0$. We need to find two numbers that add to $5$ and also multiply together to give $6$. We know that $2·3=6$ and that $2+3=5$, so the two numbers for $y$ and $z$ must be $2$ and $3$. And that’s how we factor our equation:

$$x^2+5x+6=(x+2)(x+3)$$

And now we can rewrite the entire equation as:

$$(x+2)(x+3)=0$$

Finally, we know that if two numbers multiply together to give $0$ as a product, then one of them MUST be $0$. So either $x+2=0$, in which case $x=-2$, or $x+3=0$, in which case $x=-3$. Now that we understand factoring quadratic equations, let’s get back to the question at hand.

 

So to factor $x^2-x-2=0$, we need to find two integers that multiply to $-2$ and add to $-1$. There aren’t many factors of $-2$, and we can see that $-2$ paired with $1$ satisfies both of these criteria. So let’s use $-2$ and $1$ to factor our quadratic equation.

$$\;\;\;\;\;x^2-x-2=0$$
$$(x-2)(x+1)=0$$

Now for the product of two expressions to be equal to $0$, one of the expressions MUST be equal to $0$. So either $x-2=0$, in which case x=2, or $x+1=0$, in which case $x=-1$. So the two $x$-values where the graphs intersect are $x=-1$ and $x=2$.

We can get the $y$-coordinate corresponding to these $x$-coordinates using either of the two equations for our graphs. Let’s arbitrarily use $y=x+2$. So this equation tells us that we can get the $y$-coordinates by just adding $2$ to the $x$-coordinates. Doing so, we get two points: $(-1,1)$ and $(2,4)$.

The correct answers are B and E.

What Did We Learn

Factoring quadratic equations is definitely a useful skill to master for the GRE. Let’s be sure we’re comfortable with it before test day.

 

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