Problem 63 - Beginning and Intermediate Algebra

Question

One solution of \(4 x^{2}+b x-3=0\) is \(-\frac{5}{2} .\) Find \(b\) and the other solution.

Step-by-Step Solution

Answer

b is 8.8 and the other solution is \(\frac{3}{10}\).

Step 1: Identify the known values

Given the quadratic equation is \(4x^{2} + bx - 3 = 0\) and one solution is \(-\frac{5}{2}\).

Step 2: Use Vieta's formulas

Vieta's formulas state that for a quadratic equation \(ax^2 + bx + c = 0\) the sum of the roots \(r_1 + r_2 = -\frac{b}{a}\) and the product of the roots \(r_1 \times r_2 = \frac{c}{a}\).

Step 3: Express the sum of the roots

The sum of the roots \(r_1 + r_2 = -\frac{b}{4}\). Given one root \(-\frac{5}{2}\), let the other root be \(r\). Then the sum is \(-\frac{5}{2} + r = -\frac{b}{4}\).

Step 4: Express the product of the roots

The product of the roots \(\left(-\frac{5}{2}\right) \times r = \frac{-3}{4}\).

Step 5: Solve for the other root

Solve \(\left(-\frac{5}{2}\right) \times r = \frac{-3}{4}\) for \(r\): \[ r = \frac{-3}{4} \div -\frac{5}{2} = \frac{-3}{4} \times -\frac{2}{5} = \frac{6}{20} = \frac{3}{10} \].

Step 6: Substitute to find b

Substitute \(r = \frac{3}{10}\): \[ -\frac{5}{2} + \frac{3}{10} = -\frac{b}{4} \]. Simplify the sum: \[ -\frac{25}{10} + \frac{3}{10} = -\frac{22}{10} = -\frac{11}{5} \]. Thus, \[ -\frac{11}{5} = -\frac{b}{4} \]. Solve for \(b\): \[ \frac{11}{5} = \frac{b}{4} \Rightarrow b = 4 \times \frac{11}{5} = \frac{44}{5} = 8.8 \].

Key Concepts

Vieta's formulas

Vieta's formulas are a useful tool when solving quadratic equations. They help relate the coefficients of the equation to the sums and products of its roots. For any quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's formulas state the following:

• The sum of the roots \(r_1 + r_2 = -\frac{b}{a}\)
• The product of the roots \(r_1 \times r_2 = \frac{c}{a}\)

In our example, we have the quadratic equation \(4x^2 + bx - 3 = 0\). Using Vieta's formulas:

• The sum of the roots is \(-\frac{b}{4}\)
• The product of the roots is \(\frac{-3}{4}\)

This information is crucial to find the unknown coefficients or to cross-verify the solutions.

Roots of a quadratic equation

The 'roots' of a quadratic equation are the values of \(x\) that satisfy the equation. These roots can either be real or complex numbers. Given a quadratic equation in standard form \(ax^2 + bx + c = 0\), the roots are \(r_1\) and \(r_2\).

• Real roots if the discriminant (\(b^2 - 4ac\)) is positive or zero.
• Complex roots if the discriminant is negative.

In the exercise, we already know one of the roots: \(-\frac{5}{2}\). We denote the other root as \(r\). By using Vieta's product formula, we can solve for the other root:
\left( -\frac{5}{2} \right) \times r = \frac{-3}{4} \
Simplifying this, we get \(r = \frac{3}{10}\). By establishing both roots, we are better positioned to find the remaining coefficients.

Quadratic equation solutions

Solving a quadratic equation involves finding its roots. We can do this using a variety of methods including factoring, completing the square, or applying the quadratic formula. However, Vieta's formulas provide a quick way to find missing coefficients when some roots are known.

In the given exercise, we followed these steps:

• Used Vieta's formulas to express the sum and product of the roots in terms of \(b\).
• Substituted the known root \(-\frac{5}{2}\) and solved for the other root \( \frac{3}{10} \).
• With both roots known, used the sum of the roots to solve for \(b\) as follows:

\left(-\frac{5}{2} + \frac{3}{10} = -\frac{b}{4}\right)\
Simplifying, we find: \left(-\frac{25}{10} + \frac{3}{10} = -\frac{22}{10} = -\frac{11}{5} \right)\.
Thus, \left(b = 4 \times \frac{11}{5} = 8.8\right).

By systematically breaking down the problem and using Vieta's formulas, we derived the unknown coefficient \(b\).