Questions: 3x^2 + 10x + 3 = 0 3x^2 + 7x + 2 = 0

3x^2 + 10x + 3 = 0 3x^2 + 7x + 2 = 0
Transcript text: $3 x^{2}+10 x+3=0$ $3 x^{2}+7 x+2=0$
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Solution

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Solution Steps

Solution Approach
  1. For each quadratic equation, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots.
  2. Identify the coefficients \(a\), \(b\), and \(c\) from each equation.
  3. Substitute these coefficients into the quadratic formula to compute the solutions.
Step 1: Solve the First Equation

For the equation \(3x^2 + 10x + 3 = 0\), we apply the quadratic formula:

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

Substituting \(a = 3\), \(b = 10\), and \(c = 3\):

\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{-10 \pm \sqrt{100 - 36}}{6} = \frac{-10 \pm \sqrt{64}}{6} \]

Calculating the roots:

\[ x = \frac{-10 \pm 8}{6} \]

This gives us two solutions:

\[ x_1 = \frac{-2}{6} = -\frac{1}{3}, \quad x_2 = \frac{-18}{6} = -3 \]

Step 2: Solve the Second Equation

For the equation \(3x^2 + 7x + 2 = 0\), we again use the quadratic formula:

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

Substituting \(a = 3\), \(b = 7\), and \(c = 2\):

\[ x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{-7 \pm \sqrt{49 - 24}}{6} = \frac{-7 \pm \sqrt{25}}{6} \]

Calculating the roots:

\[ x = \frac{-7 \pm 5}{6} \]

This gives us two solutions:

\[ x_1 = \frac{-2}{6} = -\frac{1}{3}, \quad x_2 = \frac{-12}{6} = -2 \]

Final Answer

The roots of the first equation \(3x^2 + 10x + 3 = 0\) are:

\[ \boxed{x = -\frac{1}{3}, -3} \]

The roots of the second equation \(3x^2 + 7x + 2 = 0\) are:

\[ \boxed{x = -\frac{1}{3}, -2} \]

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