Question

question 7 (mandatory) (1 point)
determine the roots of $3.3x^2 + 1.9x - 2.4 = 0$ to the nearest hundredth.
\\(\bigcirc\\) a) $-0.61$ and $1.19$
\\(\bigcirc\\) b) no real solution
\\(\bigcirc\\) c) $1.19$
\\(\bigcirc\\) d) $-1.19$ and $0.61$

Explanation:

Step1: Identify coefficients

For quadratic equation \(ax^2 + bx + c = 0\), here \(a = 3.3\), \(b = 1.9\), \(c = -2.4\).

Step2: Use quadratic formula

Quadratic formula: \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)

First, calculate discriminant \(D = b^2 - 4ac\)
\(D=(1.9)^2-4\times3.3\times(-2.4)\)
\(D = 3.61+31.68\)
\(D = 35.29\)

Then, find roots:
\(x=\frac{-1.9\pm\sqrt{35.29}}{2\times3.3}\)
\(\sqrt{35.29}=5.94\) (approx)
So, \(x_1=\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\) (Wait, no, wait: Wait, original equation is \(3.3x^2 + 1.9x - 2.4 = 0\), so when we plug into formula:

Wait, correction: \(x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-1.9\pm\sqrt{35.29}}{2\times3.3}\)

\(\sqrt{35.29} = 5.94\) (exactly, since \(5.94^2=35.2836\approx35.29\))

So, first root: \(\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)? No, wait, no, wait: Wait, \( -1.9 + 5.94 = 4.04\), divided by \(6.6\) is approximately \(0.61\)? But option d is -1.19 and 0.61. Wait, let's recalculate:

Wait, \(2a = 2\times3.3 = 6.6\)

\(-b = -1.9\)

So, first root: \(\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)

Second root: \(\frac{-1.9 - 5.94}{6.6}=\frac{-7.84}{6.6}\approx -1.19\)

Ah, there we go. So the roots are approximately -1.19 and 0.61, which is option d.

Answer:

d) -1.19 and 0.61