Quadratic Equations

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ACT Math › Quadratic Equations

Questions 1 - 10

Which of the following is a root of the function $f(x)=2x^2-7x-4$ ?

$x = -\frac {1}{2}$

$x = \frac{1}{2}$

$x = -4$

$x = -2$

$x = \frac{1}{4}$

Explanation

The roots of a function are the x intercepts of the function. Whenever a function passes through a point on the x-axis, the value of the function is zero. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values of x.

$f(x)=2x^2-7x-4$ $= 0$

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

Because the coefficient in front of the $x^2$ is not equal to 1, we need to multiply this coefficient by the constant, which is –4. When we mutiply 2 and –4, we get –8. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of the x term). Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We will now rewrite –7x as –8x + x.

$2x^2-7x-4=2x^2-8x+x-4=0$

We will then group the first two terms and the last two terms.

$(2x^2-8x)+(x-4)=0$

We will next factor out a 2_x_ from the first two terms.

$(2x^2-8x)+(x-4)=2x(x-4)+1(x-4)=(2x+1)(x-4)=0$

Thus, when factored, the original equation becomes (2_x_ + 1)(x – 4) = 0.

We now set each factor equal to zero and solve for x.

$2x + 1 = 0$

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

$x=-\frac{1}{2}$

Now, we set x – 4 equal to 0.

x – 4 = 0

Add 4 to both sides.

x = 4

The roots of f(x) occur where x = $-\frac{1}{2},4$ .

The answer is therefore $x = -\frac {1}{2}$ .

Which of the following is a root of the function $f(x)=2x^2-7x-4$ ?

$x = -\frac {1}{2}$

$x = \frac{1}{2}$

$x = -4$

$x = -2$

$x = \frac{1}{4}$

Explanation

$f(x)=2x^2-7x-4$ $= 0$

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

$2x^2-7x-4=2x^2-8x+x-4=0$

We will then group the first two terms and the last two terms.

$(2x^2-8x)+(x-4)=0$

We will next factor out a 2_x_ from the first two terms.

$(2x^2-8x)+(x-4)=2x(x-4)+1(x-4)=(2x+1)(x-4)=0$

Thus, when factored, the original equation becomes (2_x_ + 1)(x – 4) = 0.

We now set each factor equal to zero and solve for x.

$2x + 1 = 0$

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

$x=-\frac{1}{2}$

Now, we set x – 4 equal to 0.

x – 4 = 0

Add 4 to both sides.

x = 4

The roots of f(x) occur where x = $-\frac{1}{2},4$ .

The answer is therefore $x = -\frac {1}{2}$ .

Solve for :

Round to the nearest hundredth.

$-6\textup{ and}-3$

$\textup{No solution}$

$6\textup{ and }3$

$-4.44\textup{ and }2.41$

$4.44\textup{ and}-2.41$

Explanation

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can. Factor the quadratic expression:

Now, remember that you merely need to set each group equal to . This gives you the two values for :

; therefore

Likewise, for the other group,

Solve for :

Round to the nearest hundredth.

$-6\textup{ and}-3$

$\textup{No solution}$

$6\textup{ and }3$

$-4.44\textup{ and }2.41$

$4.44\textup{ and}-2.41$

Explanation

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can. Factor the quadratic expression:

Now, remember that you merely need to set each group equal to . This gives you the two values for :

; therefore

Likewise, for the other group,

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

Solve for :

$-6\textup{ and }\frac{-1}{3}$

$\textup{No solution}$

$3\textup{ and }6$

$\frac{2}{5}\textup{ and }\frac{4}{5}$

$\frac{1 + \sqrt{6}}{3}\textup{ and }\frac{1 - \sqrt{6}}{3}$

Explanation

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to :

Thus, simplify into:

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can, though we are sometimes a bit intimidated by terms that have a coefficient like this. Factor the quadratic expression: