Certainly! Let's solve the quadratic equation [tex]\(3x^2 - 7x + 4 = 0\)[/tex] step-by-step using the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are:
[tex]\[ a = 3, \quad b = -7, \quad c = 4 \][/tex]
1. Calculate the Discriminant:
The discriminant ([tex]\(\Delta\)[/tex]) of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-7)^2 - 4(3)(4) \][/tex]
[tex]\[ \Delta = 49 - 48 \][/tex]
[tex]\[ \Delta = 1 \][/tex]
2. Calculate the Roots:
With the discriminant ([tex]\(\Delta\)[/tex]) calculated, we can use the quadratic formula to find the roots.
For [tex]\(x_1\)[/tex]:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{-(-7) + \sqrt{1}}{2(3)} \][/tex]
[tex]\[ x_1 = \frac{7 + 1}{6} \][/tex]
[tex]\[ x_1 = \frac{8}{6} \][/tex]
[tex]\[ x_1 = \frac{4}{3} \][/tex]
For [tex]\(x_2\)[/tex]:
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x_2 = \frac{-(-7) - \sqrt{1}}{2(3)} \][/tex]
[tex]\[ x_2 = \frac{7 - 1}{6} \][/tex]
[tex]\[ x_2 = \frac{6}{6} \][/tex]
[tex]\[ x_2 = 1 \][/tex]
Therefore, the solutions to the equation [tex]\(3x^2 - 7x + 4 = 0\)[/tex] are:
[tex]\[ x = \frac{4}{3} \quad \text{and} \quad x = 1 \][/tex]
Among the given options, the correct solutions are:
[tex]\[ \boxed{\frac{4}{3}} \quad \text{and} \quad \boxed{1} \][/tex]
