To find the zeros of the quadratic equation \( y = 2x^2 + 5x - 12 \), we need to set the equation equal to zero and solve for \( x \):
\[ 2x^2 + 5x - 12 = 0 \]
We can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 5 \), and \( c = -12 \).
Now, we can plug in the values:
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Calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (5)^2 - 4(2)(-12) = 25 + 96 = 121 \]
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Now we substitute back into the quadratic formula: \[ x = \frac{-5 \pm \sqrt{121}}{2(2)} = \frac{-5 \pm 11}{4} \]
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This gives us two possible solutions: \[ x_1 = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2} \] \[ x_2 = \frac{-5 - 11}{4} = \frac{-16}{4} = -4 \]
Thus, the zeros of the graph are \( \frac{3}{2} \) and \( -4 \).
Therefore, the correct answer is:
C. \(\frac{3}{2}\) and -4
3 answers