To find the zeros of the function f(x)=2x2+13.5x+17.4 f(x) = 2x^{2} + 13.5x + 17.4 f(x)=2x2+13.5x+17.4, we first calculate the discriminant D D D using the formula:
D=b2−4ac D = b^2 - 4ac D=b2−4ac
Substituting the values a=2 a = 2 a=2, b=13.5 b = 13.5 b=13.5, and c=17.4 c = 17.4 c=17.4:
D=(13.5)2−4⋅2⋅17.4=182.25−139.2=43.05 D = (13.5)^2 - 4 \cdot 2 \cdot 17.4 = 182.25 - 139.2 = 43.05 D=(13.5)2−4⋅2⋅17.4=182.25−139.2=43.05
Since the discriminant D D D is positive (D>0 D > 0 D>0), we can find two distinct real roots using the quadratic formula:
x=−b±D2a x = \frac{{-b \pm \sqrt{D}}}{2a} x=2a−b±D
Using the positive square root for the first root:
x1=−13.5+43.052⋅2 x_1 = \frac{{-13.5 + \sqrt{43.05}}}{2 \cdot 2} x1=2⋅2−13.5+43.05
Using the negative square root for the second root:
x2=−13.5−43.052⋅2 x_2 = \frac{{-13.5 - \sqrt{43.05}}}{2 \cdot 2} x2=2⋅2−13.5−43.05
After calculating the roots, we round them to the nearest thousandths:
x1≈−1.735 x_1 \approx -1.735 x1≈−1.735 x2≈−5.015 x_2 \approx -5.015 x2≈−5.015
The zeros of the function are −1.735 \boxed{-1.735} −1.735 and −5.015 \boxed{-5.015} −5.015.
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