The function $f$ is defined for all numbers $x$ by $f(x)=x^2+x$. If $t$ is a number such that $f(2t)=30$, which two of the following could be the number $t$ ?

Indicate __two__ such numbers.

- $-5$
- $-3$
- $-1/2$
- $2$
- $5/2$

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 19 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of **Functions** 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 question mentions a **function** $f$, so it *probably* tests our **Functions** math skill. 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.

- We know that $f(x)=x^2+x$
- We know that $f(2t)=30$
- We want to choose the two possible values of $t$

## Develop a Plan

The question gives us two different equations for $f$. Let’s write them down on our paper, aligning the equal signs.

$f(x)$ | $=$ | $x^2+x$ |

$f(2t)$ | $=$ | $30$ |

Understanding functions often involves understanding **pattern matching**. It’s a useful skill to master for this test, and in life too! Looking at our two equations for the function $f$, it looks like the $2t$ matches up with the $x$ on the left side, and the $30$ matches up with the $x^2+x$. Well, in fact, that’s EXACTLY how we’re supposed to view this question. Since the equations describe the same function $(f)$, **similarly placed expressions in the equations should be equal to each other**. So from this we get $x=2t$ and $x^2+x=30$.

We want to solve for the two possibles values of $t$. If we had values for $x$, then we could definitely calculate $t$. So let’s solve the equation $x^2+x=30$ for $x$ then plug in the value of $x$ into the $x=2t$ equation.

## Solve the Question

We want to solve an equation for $x$ that has an $x^2$ term in it, thus making it a *quadratic equation*. We remember learning how to solve these types of equations in our **Solving Quadratic Equations** GRE math skill. In fact, we know that if we can factor this quadratic equation, we can solve for the value of $x$. Let’s quickly review factoring quadratic equations, if necessary, then solve for $x$.

## 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.

Now let’s factor our quadratic equation containing $x$. As we just reviewed, let’s first move everything over to the left side by subtracting $30$ from both sides.

$x^2+x$ | $=$ | $30$ |

$-30$ | $=$ | $-30$ |

$x^2+x-30$ | $=$ | $0$ |

From our **Factoring a Quadratic Equation** refresher, we know that we need to find two numbers that multiply to $-30$ and add to $1$. Let’s start by writing down all of the combinations of two integers that we can think of that multiply to $-30$.

$-30$ | $=$ | $(-1)·(30)$ |

$-30$ | $=$ | $(1)·(-30)$ |

$-30$ | $=$ | $(2)·(-15)$ |

$-30$ | $=$ | $(-2)·(15)$ |

$-30$ | $=$ | $(3)·(-10)$ |

$-30$ | $=$ | $(-3)·(10)$ |

$-30$ | $=$ | $(5)·(-6)$ |

$-30$ | $=$ | $(-5)·(6)$ |

Very nice. Looking for the pair of numbers that adds up to $1$, it looks like $-5$ and $6$ will work. Let’s find their sum just to be sure.

$$-5+6=1$$

Yes! It worked. Alright, now let’s plug in $-5$ and $6$ into our quadratic equation to write it in factored form.

$$\;\;\;x^2+x-30 = 0$$

$$(x-5)(x+6)=0$$

Next, we know that if the product of two numbers is $0$, then one of the numbers MUST be $0$. So either $x-5=0$, in which case $x=5$, OR $x+6=0$, in which case $x=-6$. **So the two possible values of $x$ are $5$ and $-6$**.

Now that we have our values for $x$, let’s solve for $t$. We have the equation $2t=x$. Solving it for $t$, we get $t=x/2$. And plugging in our two $x$-values, we get $t=5/2$ and $t=-3$. So we can see that **the correct answers are B, $-3$, and E, $5/2$**.

## What Did We Learn

Being able to solve quadratic equations quickly and efficiently through factoring is a very useful skill to master before test day. If necessary, let’s practice the factoring in this question again so that we’ll be more confident in our factoring ability.

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