Question

Determine the zero(s) of the quadratic function y=6x^2+13x+5. Use Algebra (factoring or quadratic formula) to prove your answer.

Answer 1

To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

Now we can calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5
\]
\[
= 169 - 120
\]
\[
= 49
\]

Since the discriminant is positive, we have two real and distinct solutions. Now we can substitute this back into the quadratic formula:

\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
\[
= \frac{-13 \pm 7}{12}
\]

Now we will calculate the two possible values for \( x \):

1. Using the positive root:
\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. Using the negative root:
\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

Thus, the zeros of the function are:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]

To verify these results, we can factor the quadratic \( 6x^2 + 13x + 5 \) by searching for two numbers that multiply to \( 30 \) (the product of \( 6 \) and \( 5 \)) and that add up to \( 13 \):

The two numbers are \( 3 \) and \( 10 \). We can rewrite the middle term:

\[
6x^2 + 3x + 10x + 5
\]

Now group the terms:

\[
(6x^2 + 3x) + (10x + 5)
\]
\[
= 3x(2x + 1) + 5(2x + 1)
\]
\[
= (3x + 5)(2x + 1)
\]

Setting each factor to zero gives:

1. \( 3x + 5 = 0 \) which leads to \( x = -\frac{5}{3} \)
2. \( 2x + 1 = 0 \) which leads to \( x = -\frac{1}{2} \)

Thus, the factors confirm that the zeros are indeed:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]