The quadratic equation [tex]3x^2 + 45x + 24 = 0[/tex] was solved using the quadratic formula, [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]. One solution is -14.45. What is the other solution? Round to the hundredths place.

A. -1.11
B. -0.55
C. 0.52
D. 14.45



Answer :

To find the solutions to the quadratic equation [tex]\( 3x^2 + 45x + 24 = 0 \)[/tex], we can use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

For the given quadratic equation, the coefficients are:
[tex]\[ a = 3, \quad b = 45, \quad c = 24 \][/tex]

First, we'll calculate the discriminant ([tex]\( \Delta \)[/tex]):

[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 45^2 - 4 \cdot 3 \cdot 24 \][/tex]
[tex]\[ \Delta = 2025 - 288 \][/tex]
[tex]\[ \Delta = 1737 \][/tex]

The quadratic formula provides us with two potential solutions:
[tex]\[ x = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]

We are given that one of the solutions is [tex]\( x = -14.45 \)[/tex].

To find the other solution, we need to substitute the values into the formula. Let's consider both cases:

Case 1:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]

Case 2:
[tex]\[ x = \frac{-45 - \sqrt{1737}}{6} \][/tex]

Given that one solution is [tex]\( x = -14.45 \)[/tex]:

Checking Case 1:
[tex]\[ -14.45 = \frac{-45 + \sqrt{1737}}{6} \][/tex]

Solving for [tex]\( \sqrt{1737} \)[/tex]:
[tex]\[ -14.45 \cdot 6 = -45 + \sqrt{1737} \][/tex]
[tex]\[ -86.7 = -45 + \sqrt{1737} \][/tex]
[tex]\[ \sqrt{1737} = -86.7 + 45 \][/tex]
[tex]\[ \sqrt{1737} = -41.7 \][/tex]
This does not provide a valid square root, so this case is incorrect.

Checking Case 2:
[tex]\[ x = \frac{-45 - \sqrt{1737}}{6} \][/tex]

Given [tex]\( -14.45 \neq \frac{-45 + \sqrt{1737}}{6} \)[/tex], it must be the following solution:

[tex]\[ -14.45 = \frac{-45 - \sqrt{1737}}{6} \][/tex]

To find the other solution:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]

Calculate:
[tex]\[ x = \frac{-45 + \sqrt{1737}}{6} \][/tex]
[tex]\[ x = \frac{-45 + 41.7}{6} \][/tex]
[tex]\[ x = \frac{-3.3}{6} \][/tex]
[tex]\[ x \approx -0.55 \][/tex]

Thus, the other solution is approximately [tex]\(-0.55\)[/tex].

The answer is:
[tex]\[ \boxed{-0.55} \][/tex]