Question

use the quadratic formula to determine the zeros for each function. round the solutions to the nearest hundredth.
(a) f(x)=2x² + 10x - 1.02
(b) h(x)=3x² + 11x - 2

Explanation:

Step1: Recall quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step2: Solve for zeros of $f(x)=2x^{2}+10x - 1.02$

For $f(x)=2x^{2}+10x - 1.02$, we have $a = 2$, $b=10$, and $c=-1.02$. First, calculate the discriminant $\Delta=b^{2}-4ac=(10)^{2}-4\times2\times(-1.02)=100 + 8.16 = 108.16$. Then $x=\frac{-10\pm\sqrt{108.16}}{2\times2}=\frac{-10\pm10.4}{4}$.

• $x_1=\frac{-10 + 10.4}{4}=\frac{0.4}{4}=0.1$
• $x_2=\frac{-10-10.4}{4}=\frac{-20.4}{4}=-5.1$

Step3: Solve for zeros of $h(x)=3x^{2}+11x - 2$

For $h(x)=3x^{2}+11x - 2$, we have $a = 3$, $b = 11$, and $c=-2$. Calculate the discriminant $\Delta=b^{2}-4ac=(11)^{2}-4\times3\times(-2)=121+24 = 145$. Then $x=\frac{-11\pm\sqrt{145}}{2\times3}=\frac{-11\pm12.04159}{6}$.

• $x_1=\frac{-11 + 12.04159}{6}=\frac{1.04159}{6}\approx0.17$
• $x_2=\frac{-11-12.04159}{6}=\frac{-23.04159}{6}\approx - 3.84$

Answer:

For $f(x)=2x^{2}+10x - 1.02$, the zeros are $x = 0.1$ and $x=-5.1$.
For $h(x)=3x^{2}+11x - 2$, the zeros are $x\approx0.17$ and $x\approx - 3.84$