Sure! Let's solve the quadratic equation [tex]\(2m^2 - 3m - 4 = 0\)[/tex] step-by-step using the quadratic formula.
The quadratic formula for finding the roots [tex]\(m\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by
[tex]\[
m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\][/tex]
1. Identify coefficients:
From the quadratic equation [tex]\(2m^2 - 3m - 4 = 0\)[/tex], the coefficients are:
[tex]\[
a = 2, \quad b = -3, \quad c = -4.
\][/tex]
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[
\Delta = b^2 - 4ac.
\][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = (-3)^2 - 4 \cdot 2 \cdot (-4) = 9 + 32 = 41.
\][/tex]
3. Solve for the roots using the quadratic formula:
The roots [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] are given by:
[tex]\[
m = \frac{-b \pm \sqrt{\Delta}}{2a}.
\][/tex]
Substitute [tex]\(b = -3\)[/tex], [tex]\(\Delta = 41\)[/tex], and [tex]\(a = 2\)[/tex]:
[tex]\[
m = \frac{-(-3) \pm \sqrt{41}}{2 \cdot 2} = \frac{3 \pm \sqrt{41}}{4}.
\][/tex]
4. Calculate the individual roots:
- For [tex]\(m_1\)[/tex] (the root with the positive square root):
[tex]\[
m_1 = \frac{3 + \sqrt{41}}{4} \approx \frac{3 + 6.403}{4} \approx \frac{9.403}{4} \approx 2.350781059358212.
\][/tex]
- For [tex]\(m_2\)[/tex] (the root with the negative square root):
[tex]\[
m_2 = \frac{3 - \sqrt{41}}{4} \approx \frac{3 - 6.403}{4} \approx \frac{-3.403}{4} \approx -0.8507810593582121.
\][/tex]
So, the solution to the quadratic equation [tex]\(2m^2 - 3m - 4 = 0\)[/tex] yields:
- The discriminant [tex]\(\Delta = 41\)[/tex].
- The roots are approximately [tex]\(m_1 \approx 2.350781059358212\)[/tex] and [tex]\(m_2 \approx -0.8507810593582121\)[/tex].
