Question

4x^2 - 17x + 15 9th grade algebra

Answer 1

To solve the quadratic equation \(4x^2 - 17x + 15 = 0\), you can use the quadratic formula, which is:

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

In this case, the coefficients are:
- \(a = 4\)
- \(b = -17\)
- \(c = 15\)

First, calculate the discriminant \(b^2 - 4ac\):

\[
b^2 = (-17)^2 = 289
\]
\[
4ac = 4 \cdot 4 \cdot 15 = 240
\]
\[
b^2 - 4ac = 289 - 240 = 49
\]

Now that we have the discriminant, we can substitute \(a\), \(b\), and the discriminant back into the quadratic formula:

\[
x = \frac{{-(-17) \pm \sqrt{49}}}{2 \cdot 4}
\]
\[
x = \frac{{17 \pm 7}}{8}
\]

Now calculate the two possible solutions:

1. \(x = \frac{{17 + 7}}{8} = \frac{24}{8} = 3\)
2. \(x = \frac{{17 - 7}}{8} = \frac{10}{8} = \frac{5}{4} = 1.25\)

So the solutions to the equation \(4x^2 - 17x + 15 = 0\) are:

\[
x = 3 \quad \text{and} \quad x = \frac{5}{4} \text{ or } 1.25
\]