Problem 41 - Algebra for College Students

Question

Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$6 t^{2}+t-3=0$$

Step-by-Step Solution

Answer

Roots are \(t_1 = \frac{-1 + \sqrt{73}}{12}\) and \(t_2 = \frac{-1 - \sqrt{73}}{12}\).

Step 1: Identify the coefficients

The given quadratic equation is \(6t^2 + t - 3 = 0\). Here, identify the coefficients for the quadratic formula \(ax^2 + bx + c = 0\). We have \(a = 6\), \(b = 1\), and \(c = -3\).

Step 2: Apply the quadratic formula

The quadratic formula is given by \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the identified coefficients \(a = 6\), \(b = 1\), and \(c = -3\) into this formula. We get: \[t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}\].

Step 3: Calculate the discriminant

First, calculate the discriminant \(b^2 - 4ac\). Here it is \(1 - 4 \times 6 \times (-3) = 1 + 72 = 73\).

Step 4: Solve for the roots

Now substitute the discriminant back into the quadratic formula: \(t = \frac{-1 \pm \sqrt{73}}{12}\). This gives the roots: \(t_1 = \frac{-1 + \sqrt{73}}{12}\) and \(t_2 = \frac{-1 - \sqrt{73}}{12}\).

Step 5: Verify with sum and product

Verify that the sum and product of the roots are correct. The sum of the roots is \(t_1 + t_2 = -\frac{b}{a} = -\frac{1}{6}\), and the product is \(t_1 \times t_2 = \frac{c}{a} = -\frac{1}{2}\). Calculate: \(t_1 + t_2 = \frac{-1 + \sqrt{73}}{12} + \frac{-1 - \sqrt{73}}{12} = \frac{-2}{12} = -\frac{1}{6}\); \(t_1 \times t_2 = \frac{-1 + \sqrt{73}}{12} \cdot \frac{-1 - \sqrt{73}}{12} = \frac{(-1)^2 - (\sqrt{73})^2}{144} = \frac{-72}{144} = -\frac{1}{2}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations that are in the form \(ax^2 + bx + c = 0\). It provides a way to solve any quadratic equation by computing the values of the variable that satisfy the equation. The formula is:

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

To use it, simply insert the equation's coefficients \(a\), \(b\), and \(c\) into the formula. This replaces trial and error methods or factorization which can be cumbersome. The result will give you the roots of the equation, which are the values of \(x\) that make the equation true. Remember:

• "\(b\)" is the coefficient of \(x\)
• "\(a\)" is the coefficient of \(x^2\)
• "\(c\)" is the constant term.

By applying the quadratic formula, you can solve any quadratic equation even if it cannot be factored.

Discriminant

The discriminant helps determine the nature of the roots of the quadratic equation and is an important part of the quadratic formula. It is the expression under the square root: \(b^2 - 4ac\). The value of the discriminant informs us about the types of solutions to the equation:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, which means the roots are identical.
• If it is negative, the roots are complex, indicating no real solutions exist.

For the equation \(6t^2 + t - 3 = 0\), the discriminant is calculated as \(1 - 4 \times 6 \times (-3) = 73\). Since 73 is positive, the equation has two distinct real roots.

Roots of Polynomial

Roots of a polynomial are the solutions that satisfy the polynomial equation, meaning substituting them back into the original polynomial results in zero. For quadratic equations,the roots are determined by solving the equation using methods like factoring or the quadratic formula. In our example, applying the quadratic formula to \(6t^2 + t - 3 = 0\) gives:

• \(t_1 = \frac{-1 + \sqrt{73}}{12}\)
• \(t_2 = \frac{-1 - \sqrt{73}}{12}\)

These two values are the equations' solutions or 'roots'. The roots are critical as they signify the points where the graph of the polynomial crosses the \(t\)-axis.

Sum and Product of Roots

The relationships involving the sum and product of the roots of quadratic equations provide a quick verification method for checking solutions. For any quadratic equation written as \(ax^2 + bx + c = 0\), the relationships are:

• The sum of the roots, \(t_1 + t_2 = -\frac{b}{a}\)
• The product of the roots, \(t_1 \times t_2 = \frac{c}{a}\)

For our example, the sum is \(-\frac{1}{6}\) and the product is \(-\frac{1}{2}\). Substituting in the calculated roots, we confirmed these values:

• Sum: \(\frac{-1 + \sqrt{73}}{12} + \frac{-1 - \sqrt{73}}{12} = -\frac{1}{6}\)
• Product: \(\frac{-1 + \sqrt{73}}{12} \cdot \frac{-1 - \sqrt{73}}{12} = -\frac{1}{2}\)

These checks affirm that the roots found using the quadratic formula are accurate.