Solved on Oct 24, 2023

Find the roots of the quadratic equation $2x^2 - x - 28 = 0$ by factoring.

STEP 1

Assumptions1. The quadratic function is $f(x)=x^-x-28$ . We are finding the zeros by factoring

STEP 2

To find the zeros of the quadratic function, we first need to factorize the quadratic equation. The general form of a quadratic equation is $ax^2 + bx + c =0$ . We need to find two numbers that multiply to $a \cdot c$ (product) and add up to $b$ (sum).

STEP 3

In our case, $a =2$ , $b = -1$ and $c = -28$ . So, we need to find two numbers that multiply to $2 \cdot -28 = -56$ and add up to $-1$ .

STEP 4

The two numbers that satisfy these conditions are $-8$ and $7$ because $-8 \cdot7 = -56$ and $-8 +7 = -1$ .

STEP 5

Now, we rewrite the middle term of the quadratic equation as the sum of the terms $-8x$ and $7x$ .
$f(x) =2x^2 -8x +7x -28$

STEP 6

We factor by grouping. First, group the first two terms together and the last two terms together in the equation.
$f(x) = (2x^2 -8x) + (x -28)$

STEP 7

Factor out the greatest common factor (GCF) from each group.
$f(x) =2x(x -4) +7(x -4)$

STEP 8

Now, you can see that $(x -4)$ is a common factor. Factor this out.
$f(x) = (2x +7)(x -4)$

STEP 9

Setting this equation equal to zero gives the solutions to the equation, which are the zeros of the function.
$2x +7 = \quad or \quad x -4 =$

STEP 10

olve these two equations to find the zeros.
For $2x +7 =0$ , subtract $7$ from both sides and then divide by $2$ to get $x = -\frac{7}{2}$ .
For $x -4 =0$ , add $4$ to both sides to get $x =4$ .
So, the zeros of the function $f(x)=2x^2-x-28$ are $x = -\frac{7}{2}$ and $x =4$ .

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Find the roots of the quadratic equation $2x^2 - x - 28 = 0$ by factoring.

STEP 1

Assumptions1. The quadratic function is $f(x)=x^-x-28$ . We are finding the zeros by factoring

STEP 2

To find the zeros of the quadratic function, we first need to factorize the quadratic equation. The general form of a quadratic equation is $ax^2 + bx + c =0$ . We need to find two numbers that multiply to $a \cdot c$ (product) and add up to $b$ (sum).

STEP 3

In our case, $a =2$ , $b = -1$ and $c = -28$ . So, we need to find two numbers that multiply to $2 \cdot -28 = -56$ and add up to $-1$ .

STEP 4

The two numbers that satisfy these conditions are $-8$ and $7$ because $-8 \cdot7 = -56$ and $-8 +7 = -1$ .

STEP 5

Now, we rewrite the middle term of the quadratic equation as the sum of the terms $-8x$ and $7x$ .
$f(x) =2x^2 -8x +7x -28$

STEP 6

We factor by grouping. First, group the first two terms together and the last two terms together in the equation.
$f(x) = (2x^2 -8x) + (x -28)$

STEP 7

Factor out the greatest common factor (GCF) from each group.
$f(x) =2x(x -4) +7(x -4)$

STEP 8

Now, you can see that $(x -4)$ is a common factor. Factor this out.
$f(x) = (2x +7)(x -4)$

STEP 9

Setting this equation equal to zero gives the solutions to the equation, which are the zeros of the function.
$2x +7 = \quad or \quad x -4 =$

STEP 10

olve these two equations to find the zeros.
For $2x +7 =0$ , subtract $7$ from both sides and then divide by $2$ to get $x = -\frac{7}{2}$ .
For $x -4 =0$ , add $4$ to both sides to get $x =4$ .
So, the zeros of the function $f(x)=2x^2-x-28$ are $x = -\frac{7}{2}$ and $x =4$ .

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Find the roots of the quadratic equation 2x2−x−28=02x^2 - x - 28 = 02x2−x−28=0 by factoring.

STEP 1

Assumptions1. The quadratic function is f(x)=x−x−28f(x)=x^-x-28f(x)=x−x−28 . We are finding the zeros by factoring

STEP 2

To find the zeros of the quadratic function, we first need to factorize the quadratic equation. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c =0ax2+bx+c=0. We need to find two numbers that multiply to a⋅ca \cdot ca⋅c (product) and add up to bbb (sum).

STEP 3

In our case, a=2a =2a=2, b=−1b = -1b=−1 and c=−28c = -28c=−28. So, we need to find two numbers that multiply to 2⋅−28=−562 \cdot -28 = -562⋅−28=−56 and add up to −1-1−1.

STEP 4

The two numbers that satisfy these conditions are −8-8−8 and 777 because −8⋅7=−56-8 \cdot7 = -56−8⋅7=−56 and −8+7=−1-8 +7 = -1−8+7=−1.

STEP 5

Now, we rewrite the middle term of the quadratic equation as the sum of the terms −8x-8x−8x and 7x7x7x.f(x)=2x2−8x+7x−28f(x) =2x^2 -8x +7x -28f(x)=2x2−8x+7x−28

STEP 6

We factor by grouping. First, group the first two terms together and the last two terms together in the equation.f(x)=(2x2−8x)+(x−28)f(x) = (2x^2 -8x) + (x -28)f(x)=(2x2−8x)+(x−28)

STEP 7

Factor out the greatest common factor (GCF) from each group.f(x)=2x(x−4)+7(x−4)f(x) =2x(x -4) +7(x -4)f(x)=2x(x−4)+7(x−4)

STEP 8

Now, you can see that (x−4)(x -4)(x−4) is a common factor. Factor this out.f(x)=(2x+7)(x−4)f(x) = (2x +7)(x -4)f(x)=(2x+7)(x−4)

STEP 9

Setting this equation equal to zero gives the solutions to the equation, which are the zeros of the function.2x+7=orx−4=2x +7 = \quad or \quad x -4 =2x+7=orx−4=

STEP 10

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Find the roots of the quadratic equation $2x^2 - x - 28 = 0$ by factoring.

Assumptions1. The quadratic function is $f(x)=x^-x-28$ . We are finding the zeros by factoring

To find the zeros of the quadratic function, we first need to factorize the quadratic equation. The general form of a quadratic equation is $ax^2 + bx + c =0$ . We need to find two numbers that multiply to $a \cdot c$ (product) and add up to $b$ (sum).

In our case, $a =2$ , $b = -1$ and $c = -28$ . So, we need to find two numbers that multiply to $2 \cdot -28 = -56$ and add up to $-1$ .

The two numbers that satisfy these conditions are $-8$ and $7$ because $-8 \cdot7 = -56$ and $-8 +7 = -1$ .

Now, we rewrite the middle term of the quadratic equation as the sum of the terms $-8x$ and $7x$ .
$f(x) =2x^2 -8x +7x -28$

We factor by grouping. First, group the first two terms together and the last two terms together in the equation.
$f(x) = (2x^2 -8x) + (x -28)$

Factor out the greatest common factor (GCF) from each group.
$f(x) =2x(x -4) +7(x -4)$

Now, you can see that $(x -4)$ is a common factor. Factor this out.
$f(x) = (2x +7)(x -4)$

Setting this equation equal to zero gives the solutions to the equation, which are the zeros of the function.
$2x +7 = \quad or \quad x -4 =$