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Q10. If one root of the quadratic equation [tex]2x^2 + kx - 6 = 0[/tex], find the value of [tex]k[/tex]. Also, find the other root.

Sagot :

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].
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