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Chapter 11: Problem 15

Compare the equations \(3 x^{2}-5 x+4=0\) and \(r x^{2}+5 x+s=0\) a) How are the equations alike? b) How can both equations be solved for \(x ?\)

Short Answer

Expert verified
a) Both equations are quadratic with the degree of polynomials being 2, meaning they have two roots. They differ in their coefficients, with Equation 1 having constant coefficients and Equation 2 having variable coefficients. b) To solve both equations for x, use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The solutions for Equation 1 are: \(x = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(4)}}{2(3)}\), and for Equation 2, the solutions are: \(x = \frac{-5 \pm \sqrt{(5)^2 - 4(r)(s)}}{2(r)}\).

Step by step solution

01

Identify the equations

The given quadratic equations are: \(3x^2 - 5x + 4 = 0\) (Equation 1) \(rx^2 + 5x + s = 0\) (Equation 2)
02

Compare the equations

Both equations are quadratic equations and are written in the standard form \(ax^2 + bx + c = 0\). In this case, for Equation 1, \(a = 3\), \(b = -5\), and \(c = 4\), and for Equation 2, \(a = r\), \(b = 5\), and \(c = s\). Observing the two equations, we can see that in both equations the degree of the polynomials is 2, which means that both equations have two roots (either equal roots, distinct real roots, or complex conjugate pair roots). The difference between the two equations lies in their coefficients. In Equation 1, the coefficients are constants while in Equation 2 the coefficients are variables.
03

Solving for x using the quadratic formula

To solve both equations for x, we will use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For Equation 1: x = \( \frac{5 \pm \sqrt{(-5)^2 - 4(3)(4)}}{2(3)} \) For Equation 2: x = \( \frac{-5 \pm \sqrt{(5)^2 - 4(r)(s)}}{2(r)} \) Now, both equations have been solved for x using the quadratic formula.

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