Solved on Jan 17, 2024

Solve the quadratic equation $3x^2 + 5x = 12$ for real values of x.

STEP 1

Assumptions
1. We are given a quadratic equation $3x^{2} + 5x = 12$ .
2. We need to solve for $x$ .

STEP 2

First, we need to set the quadratic equation to zero by moving all terms to one side of the equation.
$3x^{2} + 5x - 12 = 0$

STEP 3

Next, we attempt to factor the quadratic equation, if possible. We look for two numbers that multiply to $3 \times (-12) = -36$ and add to $5$ .

STEP 4

We find that the numbers $9$ and $-4$ satisfy the conditions since $9 \times (-4) = -36$ and $9 + (-4) = 5$ .

STEP 5

We rewrite the middle term of the quadratic equation using the numbers $9$ and $-4$ .
$3x^{2} + 9x - 4x - 12 = 0$

STEP 6

Next, we factor by grouping. We group the terms as follows:
$(3x^{2} + 9x) + (-4x - 12) = 0$

STEP 7

We factor out the common factors from each group.
$3x(x + 3) - 4(x + 3) = 0$

STEP 8

Since both groups contain the factor $(x + 3)$ , we can factor it out.
$(x + 3)(3x - 4) = 0$

STEP 9

Now, we use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
$x + 3 = 0 \quad \text{or} \quad 3x - 4 = 0$

STEP 10

We solve each equation for $x$ .
First, for $x + 3 = 0$ :
$x = -3$

STEP 11

Then, for $3x - 4 = 0$ :
$3x = 4$
$x = \frac{4}{3}$

STEP 12

The solutions to the quadratic equation $3x^{2} + 5x = 12$ are:
$x = -3 \quad \text{and} \quad x = \frac{4}{3}$

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Solve the quadratic equation $3x^2 + 5x = 12$ for real values of x.

STEP 1

Assumptions
1. We are given a quadratic equation $3x^{2} + 5x = 12$ .
2. We need to solve for $x$ .

STEP 2

First, we need to set the quadratic equation to zero by moving all terms to one side of the equation.
$3x^{2} + 5x - 12 = 0$

STEP 3

Next, we attempt to factor the quadratic equation, if possible. We look for two numbers that multiply to $3 \times (-12) = -36$ and add to $5$ .

STEP 4

We find that the numbers $9$ and $-4$ satisfy the conditions since $9 \times (-4) = -36$ and $9 + (-4) = 5$ .

STEP 5

We rewrite the middle term of the quadratic equation using the numbers $9$ and $-4$ .
$3x^{2} + 9x - 4x - 12 = 0$

STEP 6

Next, we factor by grouping. We group the terms as follows:
$(3x^{2} + 9x) + (-4x - 12) = 0$

STEP 7

We factor out the common factors from each group.
$3x(x + 3) - 4(x + 3) = 0$

STEP 8

Since both groups contain the factor $(x + 3)$ , we can factor it out.
$(x + 3)(3x - 4) = 0$

STEP 9

Now, we use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
$x + 3 = 0 \quad \text{or} \quad 3x - 4 = 0$

STEP 10

We solve each equation for $x$ .
First, for $x + 3 = 0$ :
$x = -3$

STEP 11

Then, for $3x - 4 = 0$ :
$3x = 4$
$x = \frac{4}{3}$

STEP 12

The solutions to the quadratic equation $3x^{2} + 5x = 12$ are:
$x = -3 \quad \text{and} \quad x = \frac{4}{3}$

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Solve the quadratic equation 3x2+5x=123x^2 + 5x = 123x2+5x=12 for real values of x.

STEP 1

Assumptions1. We are given a quadratic equation 3x2+5x=123x^{2} + 5x = 123x2+5x=12.2. We need to solve for xxx.

STEP 2

First, we need to set the quadratic equation to zero by moving all terms to one side of the equation.3x2+5x−12=03x^{2} + 5x - 12 = 03x2+5x−12=0

STEP 3

Next, we attempt to factor the quadratic equation, if possible. We look for two numbers that multiply to 3×(−12)=−363 \times (-12) = -363×(−12)=−36 and add to 555.

STEP 4

We find that the numbers 999 and −4-4−4 satisfy the conditions since 9×(−4)=−369 \times (-4) = -369×(−4)=−36 and 9+(−4)=59 + (-4) = 59+(−4)=5.

STEP 5

We rewrite the middle term of the quadratic equation using the numbers 999 and −4-4−4.3x2+9x−4x−12=03x^{2} + 9x - 4x - 12 = 03x2+9x−4x−12=0

STEP 6

Next, we factor by grouping. We group the terms as follows:(3x2+9x)+(−4x−12)=0(3x^{2} + 9x) + (-4x - 12) = 0(3x2+9x)+(−4x−12)=0

STEP 7

We factor out the common factors from each group.3x(x+3)−4(x+3)=03x(x + 3) - 4(x + 3) = 03x(x+3)−4(x+3)=0

STEP 8

Since both groups contain the factor (x+3)(x + 3)(x+3), we can factor it out.(x+3)(3x−4)=0(x + 3)(3x - 4) = 0(x+3)(3x−4)=0

STEP 9

Now, we use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.x+3=0or3x−4=0x + 3 = 0 \quad \text{or} \quad 3x - 4 = 0x+3=0or3x−4=0

STEP 10

We solve each equation for xxx.First, for x+3=0x + 3 = 0x+3=0:x=−3x = -3x=−3

STEP 11

Then, for 3x−4=03x - 4 = 03x−4=0:3x=43x = 43x=4x=43x = \frac{4}{3}x=34​

STEP 12

The solutions to the quadratic equation 3x2+5x=123x^{2} + 5x = 123x2+5x=12 are:x=−3andx=43x = -3 \quad \text{and} \quad x = \frac{4}{3}x=−3andx=34​

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Solve the quadratic equation $3x^2 + 5x = 12$ for real values of x.

Assumptions
1. We are given a quadratic equation $3x^{2} + 5x = 12$ .
2. We need to solve for $x$ .

First, we need to set the quadratic equation to zero by moving all terms to one side of the equation.
$3x^{2} + 5x - 12 = 0$

Next, we attempt to factor the quadratic equation, if possible. We look for two numbers that multiply to $3 \times (-12) = -36$ and add to $5$ .

We find that the numbers $9$ and $-4$ satisfy the conditions since $9 \times (-4) = -36$ and $9 + (-4) = 5$ .

We rewrite the middle term of the quadratic equation using the numbers $9$ and $-4$ .
$3x^{2} + 9x - 4x - 12 = 0$

Next, we factor by grouping. We group the terms as follows:
$(3x^{2} + 9x) + (-4x - 12) = 0$

We factor out the common factors from each group.
$3x(x + 3) - 4(x + 3) = 0$

Since both groups contain the factor $(x + 3)$ , we can factor it out.
$(x + 3)(3x - 4) = 0$

Now, we use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
$x + 3 = 0 \quad \text{or} \quad 3x - 4 = 0$

We solve each equation for $x$ .
First, for $x + 3 = 0$ :
$x = -3$

Then, for $3x - 4 = 0$ :
$3x = 4$
$x = \frac{4}{3}$

The solutions to the quadratic equation $3x^{2} + 5x = 12$ are:
$x = -3 \quad \text{and} \quad x = \frac{4}{3}$