Question

question 17 (mandatory) (1 point)
determine the roots of $3.3x^2 + 1.9x - 2.4 = 0$ to the nearest hundredth.
\bigcirc a) 1.19
\bigcirc b) -1.19 and 0.61
\bigcirc c) no real solution
\bigcirc d) -0.61 and 1.19

Explanation:

Step1: Identify the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the roots are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 3.3\), \(b = 1.9\), \(c=- 2.4\).

Step2: Calculate the discriminant

The discriminant \(D=b^{2}-4ac=(1.9)^{2}-4\times3.3\times(-2.4)\).
First, \((1.9)^{2}=3.61\), and \(4\times3.3\times2.4 = 31.68\) (since \(c\) is negative, \(-4ac\) becomes positive). So \(D = 3.61+31.68=35.29\).

Step3: Find the roots

Now, \(x=\frac{-1.9\pm\sqrt{35.29}}{2\times3.3}\). Since \(\sqrt{35.29} = 5.94\) (approx).
For the plus sign: \(x=\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)? Wait, no, wait, wait. Wait, \(a = 3.3\), \(b = 1.9\), \(c=-2.4\). Wait, maybe I made a sign error. Wait, \(c=-2.4\), so \(-4ac=-4\times3.3\times(-2.4)=31.68\). So discriminant is \(1.9^{2}+31.68=3.61 + 31.68 = 35.29\), square root of 35.29 is 5.94 (exactly, since \(5.94^{2}=35.2836\approx35.29\)). Then \(x=\frac{-1.9\pm5.94}{6.6}\).
First root (plus): \(\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)? No, wait, no, wait, \(2a=6.6\). Wait, \(-1.9 + 5.94 = 4.04\), \(4.04\div6.6\approx0.61\)? But the other root: \(\frac{-1.9-5.94}{6.6}=\frac{-7.84}{6.6}\approx - 1.19\). Wait, but let's check the options. Option b is - 1.19 and 0.61? Wait, no, wait, maybe I mixed up \(a\), \(b\), \(c\). Wait, the equation is \(3.3x^{2}+1.9x - 2.4=0\), so \(a = 3.3\), \(b = 1.9\), \(c=-2.4\). So the quadratic formula is correct. Wait, but let's recalculate:
\(\sqrt{35.29}=5.94\) (exact value: \(5.94\times5.94 = 35.2836\approx35.29\)). Then:
First root: \(\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)
Second root: \(\frac{-1.9 - 5.94}{6.6}=\frac{-7.84}{6.6}\approx - 1.19\)
So the roots are - 1.19 and 0.61, which is option b? Wait, no, wait, the options: option b is - 1.19 and 0.61, option d is - 0.61 and 1.19. Wait, maybe I made a mistake in the sign of \(b\). Wait, the equation is \(3.3x^{2}+1.9x - 2.4 = 0\), so \(b = 1.9\), so \(-b=-1.9\). Wait, but let's use the quadratic formula correctly. Let's compute again:

\(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-1.9\pm\sqrt{(1.9)^{2}-4\times3.3\times(-2.4)}}{2\times3.3}\)

Calculate inside the square root: \(1.9^{2}=3.61\), \(4\times3.3\times2.4 = 31.68\) (since \(c\) is negative, \(-4ac = 4\times3.3\times2.4\)), so total inside sqrt is \(3.61 + 31.68=35.29\), sqrt(35.29)=5.94 (as 5.94^2=35.2836≈35.29). Then:

First root (with +): \(\frac{-1.9 + 5.94}{6.6}=\frac{4.04}{6.6}\approx0.61\)

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

So the roots are approximately - 1.19 and 0.61, which is option b.

Answer:

b) -1.19 and 0.61