Question

Solve this quadratic equations correct to two decimal places
3A²+5A=2A²-3A+2

Answer 1

To solve the quadratic equation \(3A^2 + 5A = 2A^2 - 3A + 2\), we first rearrange the equation to set it to zero.

1. Move all terms to one side of the equation:
\[
3A^2 + 5A - 2A^2 + 3A - 2 = 0
\]

2. Combine like terms:
\[
(3A^2 - 2A^2) + (5A + 3A) - 2 = 0
\]
\[
A^2 + 8A - 2 = 0
\]

Now we have the quadratic equation:
\[
A^2 + 8A - 2 = 0
\]

We can solve this using the quadratic formula:
\[
A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = 8\), and \(c = -2\).

3. Calculate the discriminant:
\[
b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-2) = 64 + 8 = 72
\]

4. Substitute in the quadratic formula:
\[
A = \frac{-8 \pm \sqrt{72}}{2 \cdot 1} = \frac{-8 \pm 6\sqrt{2}}{2}
\]

5. Simplify:
\[
A = \frac{-8}{2} \pm \frac{6\sqrt{2}}{2}
\]
\[
A = -4 \pm 3\sqrt{2}
\]

6. Now, calculate the two possible values for \(A\):
\[
A_1 = -4 + 3\sqrt{2} \quad \text{and} \quad A_2 = -4 - 3\sqrt{2}
\]

Calculating these to two decimal places:

- Calculate \(3\sqrt{2} \approx 3 \times 1.414 = 4.242\):
- \(A_1 \approx -4 + 4.242 \approx 0.242\)
- \(A_2 \approx -4 - 4.242 \approx -8.242\)

So the solutions to the quadratic equation \(3A^2 + 5A = 2A^2 - 3A + 2\) to two decimal places are:
\[
A_1 \approx 0.24 \quad \text{and} \quad A_2 \approx -8.24
\]