Answer :
Sure, let's solve the quadratic equation [tex]\(4x^2 + 45x + 24 = 0\)[/tex] step-by-step using the quadratic formula. The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the equation [tex]\(ax^2 + bx + c = 0\)[/tex].
Given:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 45\)[/tex]
- [tex]\(c = 24\)[/tex]
First, we calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values, we get:
[tex]\[ \Delta = 45^2 - 4 \cdot 4 \cdot 24 \][/tex]
[tex]\[ \Delta = 2025 - 384 \][/tex]
[tex]\[ \Delta = 1641 \][/tex]
Now, we use the quadratic formula to find the solutions:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
First solution already given:
[tex]\[ x_1 = -10.69 \][/tex]
To find the other solution ([tex]\(x_2\)[/tex]), we proceed as follows:
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values, we get:
[tex]\[ x_2 = \frac{-45 - \sqrt{1641}}{2 \cdot 4} \][/tex]
[tex]\[ x_2 = \frac{-45 - \sqrt{1641}}{8} \][/tex]
Calculating further gives:
[tex]\[ x_2 \approx \frac{-45 - 40.5}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-85.5}{8} \][/tex]
[tex]\[ x_2 \approx -10.688 \][/tex]
Since this is very close to the already provided value of [tex]\(x_1\)[/tex], we use the positive root for verification:
[tex]\[ x_2 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
Substituting the values, we get:
[tex]\[ x_2 = \frac{-45 + \sqrt{1641}}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-45 + 40.5}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-4.5}{8} \][/tex]
[tex]\[ x_2 \approx -0.5625 \][/tex]
Rounding to the hundredths place:
[tex]\[ x_2 \approx -0.56 \][/tex]
Thus, the other solution is [tex]\(-0.56\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{-0.56} \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the equation [tex]\(ax^2 + bx + c = 0\)[/tex].
Given:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 45\)[/tex]
- [tex]\(c = 24\)[/tex]
First, we calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values, we get:
[tex]\[ \Delta = 45^2 - 4 \cdot 4 \cdot 24 \][/tex]
[tex]\[ \Delta = 2025 - 384 \][/tex]
[tex]\[ \Delta = 1641 \][/tex]
Now, we use the quadratic formula to find the solutions:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
First solution already given:
[tex]\[ x_1 = -10.69 \][/tex]
To find the other solution ([tex]\(x_2\)[/tex]), we proceed as follows:
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values, we get:
[tex]\[ x_2 = \frac{-45 - \sqrt{1641}}{2 \cdot 4} \][/tex]
[tex]\[ x_2 = \frac{-45 - \sqrt{1641}}{8} \][/tex]
Calculating further gives:
[tex]\[ x_2 \approx \frac{-45 - 40.5}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-85.5}{8} \][/tex]
[tex]\[ x_2 \approx -10.688 \][/tex]
Since this is very close to the already provided value of [tex]\(x_1\)[/tex], we use the positive root for verification:
[tex]\[ x_2 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
Substituting the values, we get:
[tex]\[ x_2 = \frac{-45 + \sqrt{1641}}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-45 + 40.5}{8} \][/tex]
[tex]\[ x_2 \approx \frac{-4.5}{8} \][/tex]
[tex]\[ x_2 \approx -0.5625 \][/tex]
Rounding to the hundredths place:
[tex]\[ x_2 \approx -0.56 \][/tex]
Thus, the other solution is [tex]\(-0.56\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{-0.56} \][/tex]
