Problem 38 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ t^{2}+4 t+11=0 $$

Step-by-Step Solution

Answer

The solutions are \(t = -2 + i\sqrt{7}\) and \(t = -2 - i\sqrt{7}\).

Step 1: Identify the coefficients

For the quadratic equation in the form \(at^{2} + bt + c = 0\), identify the coefficients: \(a = 1\), \(b = 4\), and \(c = 11\).

Step 2: Write down the quadratic formula

The quadratic formula is given as: \[t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\].

Step 3: Substitute the coefficients into the formula

Substitute \(a = 1\), \(b = 4\), and \(c = 11\) into the quadratic formula: \[t = \frac{-4 \pm \sqrt{4^{2} - 4(1)(11)}}{2(1)}\].

Step 4: Simplify under the square root

Calculate the expression under the square root: \[b^{2} - 4ac = 4^{2} - 4(1)(11) = 16 - 44 = -28\].

Step 5: Simplify the square root of a negative number

Since the expression under the square root is negative, use the imaginary unit \(i\): \[\sqrt{-28} = \sqrt{28}i\].

Step 6: Simplify further

Recognize that \(\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}\), therefore: \[\sqrt{-28} = 2i\sqrt{7}\].

Step 7: Substitute back into the quadratic formula

Substitute \(\sqrt{-28} = 2i\sqrt{7}\) back into the quadratic formula: \[t = \frac{-4 \pm 2i\sqrt{7}}{2}\].

Step 8: Simplify the final expression

Divide the numerator by 2: \[t = -2 \pm i\sqrt{7}\].

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). This formula allows us to find the roots of the quadratic equation, which are the values of \(x\) that satisfy the equation. The quadratic formula looks like this: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
To use this formula, we simply need to identify the coefficients \(a\), \(b\), and \(c\) from our quadratic equation. Then we substitute these values into the formula.
For example, in the equation \(t^2 + 4t + 11 = 0\), the coefficients are \(a = 1\), \(b = 4\), and \(c = 11\). Substituting these into the quadratic formula, we get:
\[t = \frac{-4 \pm \sqrt{4^2 - 4(1)(11)}}{2(1)}\].

Complex Numbers

A complex number is a number that has both a real part and an imaginary part. It's written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
In our quadratic equation example, we encounter complex numbers when the value under the square root (the discriminant) is negative. For instance, when we calculated \(4^2 - 4(1)(11) = -28\), we were left with a negative number under the square root. To proceed, we need to understand how to handle the square root of negative numbers.

Imaginary Numbers

Imaginary numbers come into play when we take the square root of a negative number. The imaginary unit \(i\) is defined as \(\sqrt{-1}\). So for any negative number -b, \(\sqrt{-b} = i\sqrt{b}\).
In our example, \(\sqrt{-28} = \sqrt{28}i\). We then simplify \(\sqrt{28}\) to \(2\sqrt{7}\), giving us \(\sqrt{-28} = 2i\sqrt{7}\).
Understanding imaginary numbers is essential for solving quadratic equations where the discriminant is negative and leads to non-real solutions.

Discriminant

The discriminant \(D\) in a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\).
The discriminant helps determine the nature of the roots of the quadratic equation:

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), the equation has exactly one real root (a repeated root).
• If \(D < 0\), the equation has two complex conjugate roots.

In our example, the discriminant \(D = 4^2 - 4(1)(11) = -28\) is less than zero, indicating that the equation has complex solutions. This results in the solutions \(t = -2 \pm i\sqrt{7}\).