Solving Quadratic Functions

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SAT Subject Test in Math I › Solving Quadratic Functions

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Find the roots of the function:
$\small f(x)=x^2+4x-12$

$\small \small x=2\ or\ -6$

CORRECT

$\small x=-4\ or\ 3$

$\small x=-3\ or\ 4$

$\small \small x=-2\ or\ 6$

Explanation

Factor:

$\small \small 0=(x-2)(x+6)$

Double check by factoring:

$\small F: (x)(x)$

$\small \small O:(x)(6)=6x$

$\small \small I: (-2)(x)=-2x$

$\small \small L: (-2)(6)=-12$

Add together: $\small x^2+6x-2x-12=x^2+4x-12$

Therefore:

$\small x-2=0\ or x+6=0$

$\small x=2\or-6$

A baseball is thrown straight up with an initial speed of 50 feet per second by a man standing on the roof of a 120-foot high building. The height of the baseball in feet, as a function of time in seconds , is modeled by the function

$h(t) = -16t^{2}+50t + 120$

To the nearest tenth of a second, how long does it take for the baseball to hit the ground?

$4.7\textrm{ sec}$

CORRECT

$3.1\textrm{ sec}$

$1.6\textrm{ sec}$

$7.5 \textrm{ sec}$

$6.4\textrm{ sec}$

Explanation

When the baseball hits the ground, the height is 0, so we set $h\left ( t\right ) = 0$ . and solve for .

$-16t^{2}+50t + 120 = 0$

This can be done using the quadratic formula:

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

Set :

$t = \frac{-50 \pm \sqrt{50^{2}-4(-16)120}}{2(-16)}$

$t = \frac{-50 \pm \sqrt{2,500- (- 7,680)}}{-32}$

$t = \frac{-50 \pm \sqrt{2,500+7,680}}{-32}$

$t = \frac{-50 \pm \sqrt{10,180}}{-32}$

$t \approx \frac{-50 \pm 100.9}{-32}$

One possible solution:

$t \approx \frac{-50 +100.9}{-32} \approx -1.6$

We throw this out, since time must be positive.

The other:

$t \approx \frac{-50 -100.9}{-32} \approx 4.7$

This solution, we keep. The baseball hits the ground in about 4.7 seconds.

Solve:

$(-\frac{4}{3}, -1)$

CORRECT

$(-\infty, -2)$

$(-2, -\frac{4}{3})$

$(-\frac{4}{3}, \infty)$

Explanation

Start by changing the less than sign to an equal sign and solve for .

$x=-\frac{4}{3}, x=-1$

Now, plot these two numbers on a number line.

Notice how the number line is divided into three regions:

$(-\infty, -\frac{4}{3}), (-\frac{4}{3}, -1), (-1, \infty)$

Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.

For $(-\infty, -\frac{4}{3})$ , let

Since this number is not less than zero, the solution cannot be found in this region.

For $(-\frac{4}{3}, -1)$ , let

Since this number is less than zero, the solution can be found in this region.

For $(-1, \infty)$ let .

Since this number is not less than zero, the solution cannot be found in this region.

Because the solution is only negative in the interval $(-\frac{4}{3}, -1)$ , that must be the solution.

Define $f (x) = x^{2} - 5$ and .

Find $(g \circ f) (x)$

$(g \circ f) (x) = 2 x^{2}-3$

CORRECT

$(g \circ f) (x) =4 x ^{2} + 28 x +44$

$(g \circ f) (x) = 2x^{3} +7x^{2} - 10x - 35$

$(g \circ f) (x) =4 x ^{2} + 28 x + 54$

$(g \circ f) (x) = 2 x^{2}-17$

Explanation

By definition, $(g \circ f) (x) = g (f(x))$ , so

$(g \circ f) (x) = g ( x^{2} - 5 )$

$= 2 (x^{2}-5 ) + 7$

$= 2 x^{2}-10 + 7$

$= 2 x^{2}-3$

Find the roots for

CORRECT

Explanation

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

$x=\frac{10}{5}$ $x=\frac{-10}{5}$

Find the roots of .

$x=\frac{5}{9}, \frac{-5}{9}$

CORRECT

$x=\frac{9}{5}, \frac{-9}{5}$

Explanation

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfecct square factors of this term are and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

$x=\frac{5}{9}$ $x=\frac{-5}{9}$

Find the roots for

$x=\frac{7}{6}, \frac{-7}{6}$

CORRECT

$x=\frac{6}{7}, \frac{-6}{7}$

Explanation

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and .

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

$x=\frac{7}{6}$ $x=\frac{-7}{6}$

FInd the roots for

CORRECT

Explanation

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

$x=\frac{12}{4}$ $x=\frac{-12}{4}$

Solve for x.

$x^{2}-7x+10=0$

x = 5, 2

CORRECT

x = –4, –3

x = –5, –2

x = 4, 3

x = 5

Explanation

Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

$x^{2}-2x-5x+10=0$

Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

Now pull out the common factor, the "(x-2)," from both terms.

Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

Solve for x.

$x^{2}-5x+10=-13x-6$

x = –4

CORRECT

x = –4, 4

x = 5, 2

x = 2

x = –5, –2

Explanation

$x^{2}-5x+9=-13x-7$

First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

$x^{2}+8x+16=0$

There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

$x^{2}+4x+4x+16=0$

Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

Pull out the "(x+4)" to wind up with:

Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., $px^{2}+qx+k=0$ ), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses.