Question

section 3.5 homework

1. solve the equation by graphing.

6x² + 2x - 3 = -3x² + x - 4.
options:
a. x = -24
b. x ≈ -9.47
c. x = 4
d. x = 8
e. x ≈ -10.06
f. x = 12
g. x ≈ 6.28
h. no real solution

Explanation:

Step1: Rearrange the equation

First, we want to set the equation \(6x^{2}+2x - 3=-3x^{2}+x - 4\) to the form \(ax^{2}+bx + c = 0\).
Add \(3x^{2}-x + 4\) to both sides:
\(6x^{2}+3x^{2}+2x - x-3 + 4=0\)
Simplify to get \(9x^{2}+x + 1 = 0\)

Step2: Analyze the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 9\), \(b = 1\), \(c = 1\)), the discriminant is given by \(\Delta=b^{2}-4ac\).
Calculate \(\Delta=(1)^{2}-4\times9\times1=1 - 36=- 35\)

Step3: Determine the nature of solutions

Since \(\Delta=-35<0\), the quadratic equation \(9x^{2}+x + 1 = 0\) has no real solutions. This means the graphs of \(y = 6x^{2}+2x - 3\) and \(y=-3x^{2}+x - 4\) do not intersect (or have no real intersection points), so the original equation has no real solution.

Answer:

h. no real solution