Problem 104 - Contemporary Precalculus

Question

Suppose \(a, b, c\) are fixed real numbers such that \(b^{2}-4 a c \geq 0 .\) Let \(r\) and \(s\) be the solutions of $$ a x^{2}+b x+c=0 $$ (a) Use the quadratic formula to show that \(r+s=-b / a\) and \(r s=c / a\) (b) Use part (a) to verify that \(a x^{2}+b x+c=\) \(a(x-r)(x-s)\) (c) Use part (b) to factor \(x^{2}-2 x-1\) and \(5 x^{2}+8 x+2\)

Step-by-Step Solution

Answer

Answer: The factored form of the quadratic equation \(5x^2 + 8x + 2\) is \(5(x-r)(x-s)\).

Step 1: Calculate the roots using the quadratic formula.

The quadratic equation given is: $$ a x^{2}+b x+c=0 $$ To find its roots, we can use the quadratic formula, which is: $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. $$ Let the roots be \(r\) and \(s\), so we have the two solutions: $$ r = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad\text{and}\quad s = \frac{-b - \sqrt{b^2 - 4ac}}{2a} $$

Step 2: Prove that \(r+s=-b/a\) and \(r s=c/a\).

To show that \(r+s=-b/a\), we can add both root equations: $$ r + s = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}. $$ To show that \(r s = c/a\), first, by multiplying both root equations, we get: $$ r s = \left(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\right) $$ Now use the difference of squares formula \(\left(a^2 - b^2 = (a+b)(a-b)\right)\) and simplify: $$ r s = \frac{(-b)^2 - \left(\sqrt{b^2 - 4ac}\right)^2}{(2a)^2} = \frac{b^2 - b^2 + 4ac}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}. $$ (b)

Step 3: Verify that \(a(x^2) + b(x) + c = a(x-r)(x-s)\).

Let's prove that the left side is the same as the right side: $$ a(x-r)(x-s) = a\left(x^2 - (r+s)x + rs\right) $$ From part (a) we found that \(r+s = -b/a\) and \(rs = c/a\). Replace these expressions in the equation and simplify: $$ a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = ax^2 + bx + c $$ Now we can see that both sides match, so the equation is true. (c)

Step 4: Factor \(x^2 - 2x - 1\).

In this case, \(a = 1, b = -2,\) and \(c = -1\). We can use the values of \(r+s\) and \(rs\) from part (a): \(r+s=-\frac{b}{a}=2\) and \(rs=\frac{c}{a}=-1\). The quadratic equation in factored form is: $$ x^2 - 2x - 1 = (x-r)(x-s) $$

Step 5: Factor \(5x^2 + 8x + 2\).

In this case, \(a = 5, b = 8,\) and \(c = 2\). We can use the values of \(r+s\) and \(rs\) from part (a): \(r+s=-\frac{b}{a}=-\frac{8}{5}\) and \(rs=\frac{c}{a}=\frac{2}{5}\). The quadratic equation in factored form is: $$ 5x^2 + 8x + 2 = 5(x-r)(x-s) $$

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation. A quadratic equation is typically in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the unknown.

To find the solutions to this equation, you can use the quadratic formula:

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

This formula gives you the two potential roots, \(r\) and \(s\), of the quadratic:

• \(r = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\)
• \(s = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\)

To use this formula effectively, ensure that the discriminant \(b^2 - 4ac\) is non-negative, which guarantees real roots. The formula derives from completing the square on the general quadratic equation and ensures an exact solution for \(x\). Keep in mind that the "±" symbol indicates two solutions, as the square root operation can result in both a positive and a negative number.

factoring quadratics

Factoring quadratics is another method to find the roots of a quadratic equation. It involves expressing the quadratic in the form \(a(x-r)(x-s)\), where \(a\) is the coefficient of \(x^2\) from the original equation and \(r\) and \(s\) are the roots.

The factoring process involves rewriting the original quadratic equation based on its roots. For example, if the roots are \(r\) and \(s\), then:

• \(ax^2 + bx + c = a(x-r)(x-s)\)

To factor effectively, you'd start by determining the roots, usually through the quadratic formula first if they are not easily recognizable integers. Once the roots are known or reasonably estimated, plug them into the factored form. The terms \((x-r)\) and \((x-s)\) represent binomials whose product gives the original quadratic expression.

Factoring is useful because it not only confirms the solutions found using other methods like the quadratic formula but also provides a clear visual of where the roots intersect on a graph—specifically, where the quadratic function crosses the x-axis.

polynomial roots

Polynomial roots are the solutions of a polynomial equation where the polynomial evaluates to zero. They are essentially the values of \(x\) that make the equation true when substituted back into the original polynomial.

In the context of quadratics, these roots can be real or complex numbers depending on the discriminant, \(b^2 - 4ac\). Each quadratic can have:

• Two distinct real roots if \(b^2 - 4ac > 0\)
• One real root or a repeated root if \(b^2 - 4ac = 0\)
• Two complex roots if \(b^2 - 4ac < 0\)

Understanding the nature of the polynomial roots helps to graph the function and anticipate its behavior. For instance, real roots indicate intersecting points with the x-axis, whereas complex roots suggest that the graph does not touch or cross the x-axis.

The roots are foundational in algebra and calculus, providing insights into the polynomial's factors and the shape of its graph.

sum and product of roots

An interesting property of quadratic equations is the relationship between the coefficients of the equation and the sum and product of its roots.

The sum of the roots, for a quadratic equation \(ax^2 + bx + c = 0\), can be found using:

• \(r + s = -\frac{b}{a}\)

Meanwhile, the product of the roots is calculated as:

• \(r \cdot s = \frac{c}{a}\)

These relationships arise from expanding the factored form \(a(x-r)(x-s)\) and comparing it to the general quadratic form \(ax^2 + bx + c\). Matching coefficients allows us to derive these formulas, showcasing a beautiful symmetry in polynomial equations.

Utilizing the sum and product of roots can simplify further analysis of quadratic equations, help with factoring, and offer insight into the quadratic's properties without directly solving for the roots using the quadratic formula.