Problem 70 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$b^{4}+13 b^{2}+36=0$$

Step-by-Step Solution

Answer

The solutions are \( b = 2i, -2i, 3i, -3i \).

Step 1 - Substitute a variable

Let \( x = b^2 \). This means \( b^4 = x^2 \). The equation now becomes \( x^2 + 13x + 36 = 0 \).

Step 2 - Solve the quadratic equation

Solve the quadratic equation \( x^2 + 13x + 36 = 0 \) using the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) where \( a = 1 \), \( b = 13 \), and \( c = 36 \).

Step 3 - Apply the quadratic formula

Substitute the values into the quadratic formula: \( x = \frac{{-13 \pm \sqrt{{13^2 - 4 \times 1 \times 36}}}}{2 \times 1} \). This simplifies to \( x = \frac{{-13 \pm \sqrt{{169 - 144}}}}{2} \).

Step 4 - Simplify the square root

Further simplify inside the square root: \( x = \frac{{-13 \pm \sqrt{25}}}{2} \). Therefore, \( x = \frac{{-13 \pm 5}}{2} \).

Step 5 - Find the solutions for x

This gives us two solutions: \( x = \frac{{-13 + 5}}{2} = -4 \) and \( x = \frac{{-13 - 5}}{2} = -9 \).

Step 6 - Back-substitute for b

Recall that \( x = b^2 \). Therefore, \( b^2 = -4 \) and \( b^2 = -9 \).

Step 7 - Solve for b

Take the square root of both sides: For \( b^2 = -4 \), \( b = \pm 2i \). For \( b^2 = -9 \), \( b = \pm 3i \). Therefore, the solutions for \( b \) are \( b = 2i, -2i, 3i, -3i \).

Key Concepts

quadratic formula

The quadratic formula is a tool that helps us solve quadratic equations of the form \[ax^2 + bx + c = 0 \] The formula is written as \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] Here,

• a, b, and c are constants.
• \(\pm\) indicates we get two solutions: one with addition and one with subtraction.

The discriminant \[b^2 - 4ac\](Greek letters in LaTeX) determines the type of solutions we get. If it is positive, we have two real solutions. If it is zero, we have one real solution, and if it is negative, we get imaginary solutions.

In our exercise, the equation \[x^2 + 13x + 36 = 0\] was obtained by substituting b^2 for x. After applying the quadratic formula, the discriminant is \[169 - 144 = 25\], which is positive. Hence, we get two real solutions for x, which we then use to find the imaginary solutions for b.

imaginary solutions

Imaginary solutions come into play when we solve equations involving negative numbers under a square root. Normally, we can't take the square root of a negative number in the real number system. But with imaginary numbers, we can.

• The most basic imaginary unit is denoted as i, where \[i = \sqrt{-1}\]
• This means that i^2 = -1

In our step-by-step solution, we found that b^2 could be -4 or -9. For b^2 = -4, taking the square root on both sides gives us \[b = \pm 2i\]and for b^2 = -9, we get \[b = \pm 3i\]Hence, the imaginary solutions are b = 2i, -2i, 3i, -3i.

Imaginary solutions extend our understanding of numbers beyond the real number line, giving us a way to solve a wider variety of equations.

polynomial equations

Polynomial equations involve variables raised to whole number powers. The general form is \[a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0 \] Here,

• a_n, a_{n-1}, ..., a_1, a_0 are constants.
• The highest power of the variable (n) determines the degree of the polynomial.

Our original equation \[b^4 + 13b^2 + 36 = 0\] is a polynomial of degree 4 (quartic polynomial). By substituting b^2 = x, we reduced it to a quadratic equation, making it simpler to solve.

Polynomial equations often require techniques like substitution or factoring to break them down into more manageable forms. This not only simplifies the solving process but also helps us understand the structure and solutions of the polynomial.