Certainly! Let's solve the problem step-by-step.
1. Identify the given information:
We are given one root of the quadratic equation [tex]\(2x^2 + kx - 6 = 0\)[/tex]. Let's assume this root is [tex]\(x = 2\)[/tex].
2. Substitute the given root into the equation:
Substituting [tex]\(x = 2\)[/tex] into the equation:
[tex]\[
2(2)^2 + k(2) - 6 = 0
\][/tex]
Simplify this equation:
[tex]\[
8 + 2k - 6 = 0
\][/tex]
[tex]\[
2 + 2k = 0
\][/tex]
3. Solve for [tex]\(k\)[/tex]:
[tex]\[
2k = -2
\][/tex]
[tex]\[
k = -1
\][/tex]
So, the value of [tex]\(k\)[/tex] is [tex]\(-1\)[/tex].
4. Substitute [tex]\(k\)[/tex] back into the original quadratic equation:
Now the equation is:
[tex]\[
2x^2 - x - 6 = 0
\][/tex]
5. Find the other root of the quadratic equation:
To find the roots of [tex]\(2x^2 - x - 6 = 0\)[/tex], we use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -6\)[/tex].
Compute the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = (-1)^2 - 4(2)(-6) = 1 + 48 = 49
\][/tex]
Find the roots using the quadratic formula:
[tex]\[
x = \frac{-(-1) \pm \sqrt{49}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{1 \pm 7}{4}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{1 + 7}{4} = \frac{8}{4} = 2
\][/tex]
and
[tex]\[
x = \frac{1 - 7}{4} = \frac{-6}{4} = -\frac{3}{2}
\][/tex]
6. Conclusion:
The value of [tex]\(k\)[/tex] is [tex]\(-1\)[/tex]. The other root of the quadratic equation [tex]\(2x^2 + kx - 6 = 0\)[/tex] is [tex]\(-\frac{3}{2}\)[/tex]. The roots are [tex]\(2\)[/tex] and [tex]\(-\frac{3}{2}\)[/tex].
