SAT Math - How to use the quadratic function

The length of a rectangular piece of land is two feet more than three times its width. If the area of the land is , what is the width of that piece of land?

$\frac{-1+2\sqrt{94}}{3}:ft^2$

$\frac{-1-2\sqrt{94}}{3}:ft^2$

$\frac{1-2\sqrt{94}}{3}:ft^2$

$\frac{-1+4\sqrt{94}}{3}:ft^2$

$\frac{2-4\sqrt{94}}{6}:ft^2$

Explanation

The area of a rectangle is the product of its length by its width, which we know to be equal to in our problem. We also know that the length is equal to , where represents the width of the land. Therefore, we can write the following equation:

Distributing the outside the parentheses, we get:

$3w^{2}+2w=125$

Subtracting from each side of the equation, we get:

$3w^{2}+2w-125=0$

We get a quadratic equation, and since there is no factor of and that adds up to , we use the quadratic formula to solve this equation.

$w=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

We can first calculate the discriminant (i.e. the part under the square root)

$b^{2}-4ac= 2^{2}-4\times3 \times(-125)$

$4-4\times3\times(-125)$

$4-12\times(-125)$

We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):

$w=\frac{-2+\sqrt{1504}}{6}$ $w=\frac{-2+\sqrt{1504}}{6}$

Breaking down the square root:

$w=\frac{-2+\sqrt{2\times2\times2\times2\times2\times 47}}{6}$ $w=\frac{-2-\sqrt{2\times2\times2\times2\times2\times 47}}{6}$

We can pull two of the twos out of the square root and place a outside of it:

$w=\frac{-2+4\sqrt{2\times 47}}{6}$ $w=\frac{-2-4\sqrt{2\times 47}}{6}$

We can then multiply the and the :

$w=\frac{-2+4\sqrt{94}}{6}$ $w=\frac{-2-4\sqrt{94}}{6}$

At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of :

$w=\frac{-1+2\sqrt{94}}{3}$ $w=\frac{-1+2\sqrt{94}}{3}$

Since width is a positive value, the answer is:

$w=\frac{-1+2\sqrt{94}}{3}$

$w=\frac{-1+2\times9.695}{3}$

$w=\frac{-1+19.39}{3}$

$w=\frac{18.39}{3}=6.13$

The width of the piece of land is approximately .

If $x^{2}+5x-3=-9$ , which pair of values for are correct?

Explanation

First, set the quadratic function equal to :

$x^{2}+5x-3=-9$

$x^{2}+5x+6=0$

Then, reduce the function to its two factors:

Since one of the factors on the left hand side of the equation must equal in order for the above equation to be true,

Solving for both, we get .

How many x-intercepts does the following equation have?

Explanation

To figure out how many x-intercepts there are, we need to set out equation equal to zero.

Factor an x out

Factor inside of the parenthesis.

We can see there are 3 x-intercepts at .

If $x^{2}+x-50=-8$ , which two values of are correct?

Explanation

First, set the quadratic function equal to :

$x^{2}+x-50=-8$

$x^{2}+x-42=0$

Then, separate the function into its two component factors:

It follows from this equation that either or ,

If $x^{2}-12x+36=9$ , which two values of are correct?

Explanation

First, we set the quadratic function equal to :

$x^{2}-12x+36=9$

$x^{2}-12x+27=0$

Reduce the function to its two component factors:

Therefore, since either or ,

If x2 + 2x - 1 = 7, which answers for x are correct?

x = -4, x = 2

x = 8, x = 0

x = -4, x = -2

x = -3, x = 4

x = -5, x = 1

Explanation

x2 + 2x - 1 = 7

x2 + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

Which of the following quadratic equations has a vertex located at $\dpi{100} (3,4)$ ?

$f(x)=-2x^2+12x-14$

$f(x)=-2x^2+12x-12$

$f(x)=-2x^2+8x-2$

$f(x)=-2x^2-12x+4$

$f(x)=-2x^2-12x+58$

Explanation

The vertex form of a parabola is given by the equation:

$f(x)=a(x-h)^2 +k$ , where the point $\dpi{100} (h,k)$ is the vertex, and $\dpi{100} a$ is a constant.

We are told that the vertex must occur at $\dpi{100} (3,4)$ , so let's plug this information into the vertex form of the equation. $\dpi{100} h$ will be 3, and $\dpi{100} k$ will be 4.

$f(x)=a(x-3)^2 +4$

Let's now expand $(x-3)^2$ by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

$(x-3)^2 = (x-3)(x-3)=x^2-3x-3x+9=x^2-6x+9$

We can replace $(x-3)^2$ with $x^2-6x+9$ .

$f(x)=a(x-3)^2+4=a(x^2-6x+9)+4$

Next, distribute the $\dpi{100} a$ .

$a(x^2-6x+9)+4 = ax^2 -6ax+9a+4$

Notice that in all of our answer choices, the first term is $-2x^2$ . If we let $\dpi{100} a=-2$ , then we would have $-2x^2$ in our equation. Let's see what happens when we substitute $\dpi{100} -2$ for $\dpi{100} a$ .

$f(x)=ax^2-6ax+9a+4=(-2)x^2-6(-2)x+9(-2)+4$

$=-2x^2+12x-18+4$

To use a computer at a computer lab, Tonya must pay $0.50 per hour. She spends hours at the computer lab, and pays a total amount of , where . If this equation is graphed, what does the y-intercept represent?

The initial cost of using a computer.

The number of hours Tonya spent at the computer lab.

The total amount that Tonya paid.

The amount per hour that it costs to use a computer.

The amount per day that it costs to use a computer.

Explanation

If the equation

were graphed, the result would be a line with a slope of 0.5 and a y-intercept of 1.2. 0.5 represents the amount Tonya pays per hour to work at a computer, because we know that the rate is $0.50, and C represents the total amount she pays. Because we must add 1.2 (or $1.20) to the total amount of hours multiplied by the hourly rate in order to get the total amount Tonya paid, we can assume there is another charge for renting a computer, in addition to the hourly rate, and that that charge is $1.20. Therefore, the y-intercept of the graph of the equation represents the initial cost of using a computer.

What is the focus of the above quadratic equation?

$\left(3, \frac{41}{8}\right)$

$\left(3, 5\right)$

$\left(3, \frac{1}{8}\right)$

$\left(-3, \frac{41}{8}\right)$

$\left(-3, -\frac{41}{8}\right)$

Explanation

The focus is solved by using the following formula , where are the coordinates of the vertex, and is the distance from the vertex. To solve for , we use