Quadratic Equations

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PSAT Math › Quadratic Equations

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

Find all possible zeros for the following function.

Explanation

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

$\x-1=0\x-1+1=+1\x=1$

and

$\x+1=0\x+1-1=-1\x=-1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

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

Find all possible zeros for the following function.

Explanation

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

$\x-1=0\x-1+1=+1\x=1$

and

$\x+1=0\x+1-1=-1\x=-1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

Find the zeros of the following function.

Explanation

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-3, or b in the standard quadratic formula). Because their product is positive (2) and the sum is negative, that must mean that they both have negative signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 2, as the product of 1 and 2 is 2, and sum of 1 and 2 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

$\x-2=0 \x-2+2=+2 \x=2$

and

$\x-1=0 \x-1+1=+1 \x=1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

Find the zeros of the following function.

Explanation

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-3, or b in the standard quadratic formula). Because their product is positive (2) and the sum is negative, that must mean that they both have negative signs.

From here, set each binomial equal to zero and solve for .

$\x-2=0 \x-2+2=+2 \x=2$

and

$\x-1=0 \x-1+1=+1 \x=1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

Complete the square to calculate the maximum or minimum point of the given function.

Explanation

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 4$

Now divide the middle term coefficient by two.

$\frac{4}{2}=2$

From here write the function with the perfect square.

$-(x^2+4x+1)\Rightarrow-((x+2)^2+1+2^2)$

When simplified the new function is,

$\-((x+2)^2+5)$

Since the term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

$\x+2=0 \x+2-2=0-2 \x=-2$

From here, substitute the the value into the original function.

$\y=-(-2)^2-4(-2)-1 \y=-4+8-1 \y=3$

Therefore the maximum value occurs at the point .

Complete the square to calculate the maximum or minimum point of the given function.

Explanation

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

First factor out a negative one.

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 4$

Now divide the middle term coefficient by two.

$\frac{4}{2}=2$

From here write the function with the perfect square.

$-(x^2+4x+1)\Rightarrow-((x+2)^2+1+2^2)$

When simplified the new function is,

$\-((x+2)^2+5)$

$\x+2=0 \x+2-2=0-2 \x=-2$

From here, substitute the the value into the original function.

$\y=-(-2)^2-4(-2)-1 \y=-4+8-1 \y=3$

Therefore the maximum value occurs at the point .

Find all possible zeros for the following function.

Explanation

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (3, or b in the standard quadratic formula). Because their product is negative (-4) and the sum is positive, that must mean that they have opposite signs.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and -1, as the product of 4 and -1 is -4, and sum of 4 and -1 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

$\x+4=0 \x+4-4=-4 \x=-4$

and

$\x-1=0 \x-1+1=1 \x=1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

Find all possible zeros for the following function.

Explanation

To find the zeros of the function use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (3, or b in the standard quadratic formula). Because their product is negative (-4) and the sum is positive, that must mean that they have opposite signs.

From here, set each binomial equal to zero and solve for .

$\x+4=0 \x+4-4=-4 \x=-4$

and

$\x-1=0 \x-1+1=1 \x=1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

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