Problem 102 - Algebra and Trigonometry

Question

The quadratic formula gives us the roots of a quadratic equation from its coefficients. We can also obtain the coefficients from the roots. For example, find the roots of the equation \(x^{2}-9 x+20=0\) and show that the product of the roots is the constant term 20 and the sum of the roots is 9 , the negative of the coefficient of \(x .\) Show that the same relationship between roots and coefficients holds for the following equations: $$\begin{array}{l}{x^{2}-2 x-8=0} \\ {x^{2}+4 x+2=0}\end{array}$$ Use the quadratic formula to prove that in general, if the equation \(x^{2}+b x+c=0\) has roots \(r_{1}\) and \(r_{2},\) then \(c=r_{1} r_{2}\) and \(b=-\left(r_{1}+r_{2}\right)\)

Step-by-Step Solution

Answer

Roots product is constant term \( c \), sum is negative of coefficient \( b \).

Step 1: Apply the Quadratic Formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For \( x^2 - 9x + 20 = 0 \), identify \( a = 1 \), \( b = -9 \), and \( c = 20 \). Substitute these values into the quadratic formula to find the roots.

Step 2: Solve for the Roots

Calculate the roots: \( x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 80}}{2} = \frac{9 \pm 1}{2} \). Therefore, the roots are \( x = 5 \) and \( x = 4 \).

Step 3: Verify Product and Sum of the Roots

Check the product: \( 5 \times 4 = 20 \), which equals the constant term. Check the sum: \( 5 + 4 = 9 \), which is the negative of the coefficient of \( x \).

Step 4: Repeat for Second Equation

Apply the same method to \( x^2 - 2x - 8 = 0 \): Here, \( a = 1 \), \( b = -2 \), \( c = -8 \). Use the quadratic formula to find the roots.

Step 5: Calculate the Roots for Second Equation

Using the quadratic formula: \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2} \). The roots are \( x = 4 \) and \( x = -2 \).

Step 6: Verify Product and Sum for Second Equation

Check the product: \( 4 \times -2 = -8 \) (equal to the constant term \( c \)). Check the sum: \( 4 + (-2) = 2 \), matching \( -b = -(-2) = 2 \).

Step 7: Analyze Third Equation

Apply the quadratic formula to \( x^2 + 4x + 2 = 0 \): \( a = 1 \), \( b = 4 \), \( c = 2 \).

Step 8: Determine the Roots for Third Equation

Using the quadratic formula: \( x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} \). Roots are \( x = -2 + \sqrt{2} \) and \( x = -2 - \sqrt{2} \).

Step 9: Verify Product and Sum for Third Equation

Check the product: \( (-2 + \sqrt{2})(-2 - \sqrt{2}) = 4 - 2 = 2 \), (equals the constant term \( c \)). Check the sum: \( -2 + \sqrt{2} + (-2 - \sqrt{2}) = -4 \), matching \( -b = -4 \).

Step 10: Generalize the Relationship

For quadratic equation \( x^2 + bx + c = 0 \), roots \( r_1 \) and \( r_2 \) can be expressed as \( r_1 = \frac{-b + \sqrt{b^2 - 4c}}{2} \) and \( r_2 = \frac{-b - \sqrt{b^2 - 4c}}{2} \). By calculating \( r_1 \times r_2 = c \) and \( r_1 + r_2 = -b \), the relationship \( c = r_1r_2 \) and \( b = -(r_1 + r_2) \) is confirmed.

Key Concepts

Roots of Quadratic Equations

When discussing quadratic equations, one of the key components is finding their roots—these are the values of \( x \) that satisfy the equation. A general quadratic equation is written as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The roots of the equation can be found using the quadratic formula:

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

This formula is derived from the process of completing the square and provides a direct way to compute the roots when you know the coefficients.
For instance, for the equation \( x^2 - 9x + 20 = 0 \), using the quadratic formula gives roots of 4 and 5, as verified by the solutions. These roots signify the points where the graph of the quadratic equation intersects the x-axis.
Understanding the nature of roots is crucial. If the discriminant \( b^2 - 4ac \) is positive, you have two distinct real roots. If it is zero, you have exactly one real root, which is also known as a repeated or double root. Conversely, if the discriminant is negative, the equation has two complex conjugate roots.

Sum and Product of Roots

A fascinating thing about the roots \( r_1 \) and \( r_2 \) of any quadratic equation \( x^2 + bx + c = 0 \) is how their sum and product relate directly to the coefficients \( b \) and \( c \). This property stems from algebraic manipulations and can be expressed neatly without knowing the specific values of the roots.

• Sum of the roots: The sum \( r_1 + r_2 \) is equal to \( -b/a \), or simply \(-b\) if \( a = 1 \). This means that the sum is the negative of the coefficient of \( x \).
• Product of the roots: The product \( r_1 \cdot r_2 \) is \( c/a \), which becomes \( c \) if \( a = 1 \). This signifies that the product is equal to the constant term.

For example, in \( x^2 - 9x + 20 = 0 \), the roots 4 and 5 indeed add up to 9 (the negative of \( -9 \)), and their product is 20 (the constant term). This remarkable relationship helps in verifying calculations and sometimes even in constructing quadratic equations just from knowing the roots.

Solving Quadratic Equations

Solving quadratic equations efficiently involves a few strategies, with the quadratic formula being one of the most reliable methods. Here's how solving typically progresses:

• Identify coefficients: Start by determining the values of \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
• Apply the quadratic formula: Substitute these coefficients into the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots.
• Simplify the expression: Solve the expression under the square root, compute the rest and simplify to find the values of \( x \).
• Verify results: Check that the solutions satisfy the original equation and are consistent with any conditions or constraints.

This systematic approach allows finding the roots, or solutions, for any quadratic equation. Sometimes, roots might need verification through plugging back into the original equation or through a secondary check like comparing the sum and product of the roots with coefficients. This process not only solidifies confidence in the answer but also deepens understanding of the interplay between algebraic expressions and their graphical representations.