Solved on Jan 19, 2024

Find the number(s) $x$ such that $2x^2 - 12 = -5x$ .

STEP 1

Assumptions
1. Let the unknown number be represented by $x$ .
2. The phrase "twelve less than twice the square of a number" can be written as $2x^2 - 12$ .
3. The phrase "equal to the product of -5 and the number" can be written as $-5x$ .
4. We are looking for the value(s) of $x$ that satisfy the equation $2x^2 - 12 = -5x$ .

STEP 2

Write the given equation using the assumptions.
$2x^2 - 12 = -5x$

STEP 3

To solve for $x$ , we need to set the equation to zero. Move all terms to one side of the equation to form a quadratic equation.
$2x^2 + 5x - 12 = 0$

STEP 4

Now we need to solve the quadratic equation. First, we can try to factor the quadratic if possible.

STEP 5

Look for two numbers that multiply to $2 \times -12 = -24$ and add to $5$ .

STEP 6

The numbers that satisfy these conditions are $8$ and $-3$ because $8 \times -3 = -24$ and $8 + (-3) = 5$ .

STEP 7

Write the quadratic equation in its factored form using these two numbers.
$(2x - 3)(x + 4) = 0$

STEP 8

Now, we have the product of two factors equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero.

STEP 9

Set each factor equal to zero and solve for $x$ .
$2x - 3 = 0 \quad \text{or} \quad x + 4 = 0$

STEP 10

Solve the first equation $2x - 3 = 0$ for $x$ .
$2x = 3$
$x = \frac{3}{2}$

STEP 11

Solve the second equation $x + 4 = 0$ for $x$ .
$x = -4$

STEP 12

We have found two solutions for $x$ that satisfy the original equation.
The solutions are $x = \frac{3}{2}$ and $x = -4$ .

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Find the number(s) $x$ such that $2x^2 - 12 = -5x$ .

STEP 1

STEP 2

Write the given equation using the assumptions.
$2x^2 - 12 = -5x$

STEP 3

To solve for $x$ , we need to set the equation to zero. Move all terms to one side of the equation to form a quadratic equation.
$2x^2 + 5x - 12 = 0$

STEP 4

Now we need to solve the quadratic equation. First, we can try to factor the quadratic if possible.

STEP 5

Look for two numbers that multiply to $2 \times -12 = -24$ and add to $5$ .

STEP 6

The numbers that satisfy these conditions are $8$ and $-3$ because $8 \times -3 = -24$ and $8 + (-3) = 5$ .

STEP 7

Write the quadratic equation in its factored form using these two numbers.
$(2x - 3)(x + 4) = 0$

STEP 8

Now, we have the product of two factors equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero.

STEP 9

Set each factor equal to zero and solve for $x$ .
$2x - 3 = 0 \quad \text{or} \quad x + 4 = 0$

STEP 10

Solve the first equation $2x - 3 = 0$ for $x$ .
$2x = 3$
$x = \frac{3}{2}$

STEP 11

Solve the second equation $x + 4 = 0$ for $x$ .
$x = -4$

STEP 12

We have found two solutions for $x$ that satisfy the original equation.
The solutions are $x = \frac{3}{2}$ and $x = -4$ .

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Find the number(s) xxx such that 2x2−12=−5x2x^2 - 12 = -5x2x2−12=−5x.

STEP 1

STEP 2

Write the given equation using the assumptions.2x2−12=−5x 2x^2 - 12 = -5x 2x2−12=−5x

STEP 3

To solve for x x x, we need to set the equation to zero. Move all terms to one side of the equation to form a quadratic equation.2x2+5x−12=0 2x^2 + 5x - 12 = 0 2x2+5x−12=0

STEP 4

Now we need to solve the quadratic equation. First, we can try to factor the quadratic if possible.

STEP 5

Look for two numbers that multiply to 2×−12=−24 2 \times -12 = -24 2×−12=−24 and add to 5 5 5.

STEP 6

The numbers that satisfy these conditions are 8 8 8 and −3 -3 −3 because 8×−3=−24 8 \times -3 = -24 8×−3=−24 and 8+(−3)=5 8 + (-3) = 5 8+(−3)=5.

STEP 7

Write the quadratic equation in its factored form using these two numbers.(2x−3)(x+4)=0 (2x - 3)(x + 4) = 0 (2x−3)(x+4)=0

STEP 8

Now, we have the product of two factors equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero.

STEP 9

Set each factor equal to zero and solve for x x x.2x−3=0orx+4=0 2x - 3 = 0 \quad \text{or} \quad x + 4 = 0 2x−3=0orx+4=0

STEP 10

Solve the first equation 2x−3=0 2x - 3 = 0 2x−3=0 for x x x.2x=3 2x = 3 2x=3x=32 x = \frac{3}{2} x=23​

STEP 11

Solve the second equation x+4=0 x + 4 = 0 x+4=0 for x x x.x=−4 x = -4 x=−4

STEP 12

We have found two solutions for x x x that satisfy the original equation.The solutions are x=32 x = \frac{3}{2} x=23​ and x=−4 x = -4 x=−4.

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Find the number(s) $x$ such that $2x^2 - 12 = -5x$ .

Write the given equation using the assumptions.
$2x^2 - 12 = -5x$

To solve for $x$ , we need to set the equation to zero. Move all terms to one side of the equation to form a quadratic equation.
$2x^2 + 5x - 12 = 0$

Look for two numbers that multiply to $2 \times -12 = -24$ and add to $5$ .

The numbers that satisfy these conditions are $8$ and $-3$ because $8 \times -3 = -24$ and $8 + (-3) = 5$ .

Write the quadratic equation in its factored form using these two numbers.
$(2x - 3)(x + 4) = 0$

Set each factor equal to zero and solve for $x$ .
$2x - 3 = 0 \quad \text{or} \quad x + 4 = 0$

Solve the first equation $2x - 3 = 0$ for $x$ .
$2x = 3$
$x = \frac{3}{2}$

Solve the second equation $x + 4 = 0$ for $x$ .
$x = -4$

We have found two solutions for $x$ that satisfy the original equation.
The solutions are $x = \frac{3}{2}$ and $x = -4$ .