Quadratic Equations

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

Consider the equation:

$3x^{2} + 7x =$ __________

Fill in the blank with a real constant to form an equation with exactly one real solution.

$- \frac{49}{12}$

$- \frac{49}{36}$

None of the other responses gives a correct answer.

Explanation

We will call the constant that goes in the blank . The equation becomes

$3x^{2} + 7x = N$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3x^{2} + 7x - N = N - N$

$3x^{2} + 7x - N = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$7^{2} - 4(3)(-N) = 0$

Solving for :

$12N\div 12 = -49 \div 12$

$N = - \frac{49}{12}$ ,

the correct response.

Consider the equation:

$3 x^{2} + 48 =$ _________

Fill in the blank with a real constant to form an equation with exactly one real solution.

Explanation

We will call the constant that goes in the blank . The equation becomes

$3 x^{2} + 48 = Nx$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3 x^{2} + 48 - Nx = Nx - Nx$

$3 x^{2} - Nx + 48 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$(-N)^{2} - 4 (3)(48)= 0$

$N^{2} - 576= 0$

Solving for :

$N^{2} - 576+ 576 = 0+ 576$

$N^{2} = 576$

$N = \pm \sqrt{576}$

$N = \pm 24$ ,

that is, either or .

is not a choice, but 24 is; this is the correct response.

Consider the equation:

$3x^{2} + 7x =$ __________

Fill in the blank with a real constant to form an equation with exactly one real solution.

$- \frac{49}{12}$

$- \frac{49}{36}$

None of the other responses gives a correct answer.

Explanation

We will call the constant that goes in the blank . The equation becomes

$3x^{2} + 7x = N$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3x^{2} + 7x - N = N - N$

$3x^{2} + 7x - N = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$7^{2} - 4(3)(-N) = 0$

Solving for :

$12N\div 12 = -49 \div 12$

$N = - \frac{49}{12}$ ,

the correct response.

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:

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

Consider the equation:

$3 x^{2} + 48 =$ _________

Fill in the blank with a real constant to form an equation with exactly one real solution.

Explanation

We will call the constant that goes in the blank . The equation becomes

$3 x^{2} + 48 = Nx$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3 x^{2} + 48 - Nx = Nx - Nx$

$3 x^{2} - Nx + 48 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$(-N)^{2} - 4 (3)(48)= 0$

$N^{2} - 576= 0$

Solving for :

$N^{2} - 576+ 576 = 0+ 576$

$N^{2} = 576$

$N = \pm \sqrt{576}$

$N = \pm 24$ ,

that is, either or .

is not a choice, but 24 is; this is the correct response.

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 x: (x2 – x) / (x – 1) = 1

No solution

x = 1

x = -1

x = 2

x = -2

Explanation

Begin by multiplying both sides by (x – 1):

x2 – x = x – 1

Solve as a quadratic equation: x2 – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

Solve for x: (x2 – x) / (x – 1) = 1

No solution

x = 1

x = -1

x = 2

x = -2

Explanation

Begin by multiplying both sides by (x – 1):

x2 – x = x – 1

Solve as a quadratic equation: x2 – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

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