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 \( x \):
-
Using the positive root: \[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
-
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:
- \( 3x + 5 = 0 \) which leads to \( x = -\frac{5}{3} \)
- \( 2x + 1 = 0 \) which leads to \( x = -\frac{1}{2} \)
Thus, the factors confirm that the zeros are indeed: \[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
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