Math

Question Solve for yy in the quadratic equation 3y2+7y=43y^2 + 7y = -4. If there are multiple solutions, list them separated by commas. If no solution, indicate "No solution".

Studdy Solution

STEP 1

Assumptions
1. We are given a quadratic equation in the form ay2+by+c=0ay^2 + by + c = 0.
2. We need to solve for yy in the equation 3y2+7y=43y^2 + 7y = -4.

STEP 2

First, we need to rewrite the equation in the standard form of a quadratic equation, which is ay2+by+c=0ay^2 + by + c = 0.
3y2+7y+4=03y^2 + 7y + 4 = 0

STEP 3

Next, we will attempt to factor the quadratic equation, if possible. To do this, we look for two numbers that multiply to acac (where a=3a = 3 and c=4c = 4) and add up to bb (where b=7b = 7).
ac=3×4=12ac = 3 \times 4 = 12
We need two numbers that multiply to 1212 and add up to 77.

STEP 4

After examining the factors of 1212, we find that 33 and 44 are the numbers that satisfy both conditions.
3×4=123 \times 4 = 12 3+4=73 + 4 = 7

STEP 5

Now we can rewrite the middle term of the quadratic equation using the numbers 33 and 44.
3y2+3y+4y+4=03y^2 + 3y + 4y + 4 = 0

STEP 6

Next, we factor by grouping. We group the terms with common factors.
(3y2+3y)+(4y+4)=0(3y^2 + 3y) + (4y + 4) = 0

STEP 7

Factor out the common factors from each group.
3y(y+1)+4(y+1)=03y(y + 1) + 4(y + 1) = 0

STEP 8

Now, we can see that (y+1)(y + 1) is a common factor. We factor it out.
(3y+4)(y+1)=0(3y + 4)(y + 1) = 0

STEP 9

According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for yy.
3y+4=0ory+1=03y + 4 = 0 \quad \text{or} \quad y + 1 = 0

STEP 10

First, solve the equation 3y+4=03y + 4 = 0 for yy.
3y=43y = -4

STEP 11

Divide both sides by 33 to isolate yy.
y=43y = \frac{-4}{3}

STEP 12

Next, solve the equation y+1=0y + 1 = 0 for yy.
y=1y = -1
The solutions for yy are:
y=43,1y = \frac{-4}{3}, -1

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