Answer :
To solve the quadratic equation [tex]\(3x^2 + 2x - 4 = 0\)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -4\)[/tex].
Step 1: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 2^2 - 4(3)(-4) \][/tex]
[tex]\[ \Delta = 4 + 48 \][/tex]
[tex]\[ \Delta = 52 \][/tex]
Step 2: Calculate the two solutions using the quadratic formula
The solutions are given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{-2 + \sqrt{52}}{2 \cdot 3} \][/tex]
[tex]\[ x_2 = \frac{-2 - \sqrt{52}}{2 \cdot 3} \][/tex]
Step 3: Simplify and find the decimal values
First, we compute [tex]\(\sqrt{52}\)[/tex]:
[tex]\[ \sqrt{52} \approx 7.21 \][/tex]
Substitute this back into the equations for [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex]:
[tex]\[ x_1 = \frac{-2 + 7.21}{6} \][/tex]
[tex]\[ x_1 = \frac{5.21}{6} \][/tex]
[tex]\[ x_1 \approx 0.87 \][/tex]
For the second solution:
[tex]\[ x_2 = \frac{-2 - 7.21}{6} \][/tex]
[tex]\[ x_2 = \frac{-9.21}{6} \][/tex]
[tex]\[ x_2 \approx -1.54 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(3x^2 + 2x - 4 = 0\)[/tex] are:
[tex]\[ x_1 \approx 0.87 \][/tex]
[tex]\[ x_2 \approx -1.54 \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -4\)[/tex].
Step 1: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 2^2 - 4(3)(-4) \][/tex]
[tex]\[ \Delta = 4 + 48 \][/tex]
[tex]\[ \Delta = 52 \][/tex]
Step 2: Calculate the two solutions using the quadratic formula
The solutions are given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x_1 = \frac{-2 + \sqrt{52}}{2 \cdot 3} \][/tex]
[tex]\[ x_2 = \frac{-2 - \sqrt{52}}{2 \cdot 3} \][/tex]
Step 3: Simplify and find the decimal values
First, we compute [tex]\(\sqrt{52}\)[/tex]:
[tex]\[ \sqrt{52} \approx 7.21 \][/tex]
Substitute this back into the equations for [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex]:
[tex]\[ x_1 = \frac{-2 + 7.21}{6} \][/tex]
[tex]\[ x_1 = \frac{5.21}{6} \][/tex]
[tex]\[ x_1 \approx 0.87 \][/tex]
For the second solution:
[tex]\[ x_2 = \frac{-2 - 7.21}{6} \][/tex]
[tex]\[ x_2 = \frac{-9.21}{6} \][/tex]
[tex]\[ x_2 \approx -1.54 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(3x^2 + 2x - 4 = 0\)[/tex] are:
[tex]\[ x_1 \approx 0.87 \][/tex]
[tex]\[ x_2 \approx -1.54 \][/tex]
