Problem 50 - Beginning and Intermediate Algebra

Question

Find the discriminant. Use it to determine whether the solutions for each equation are A. two rational numbers B. one rational number C. two irrational numbers D. two nonreal complex numbers. Tell whether the equation can be solved using the zero-factor property, or if the quadratic formula should be used instead. Do not actually solve. $$ 9 x^{2}-12 x-1=0 $$

Step-by-Step Solution

Answer

C. Two irrational numbers; solve using the quadratic formula.

Step 1: Identify coefficients

For the quadratic equation in the form \[ax^2 + bx + c = 0\], first, identify the coefficients. Here, \(a = 9\), \(b = -12\), and \(c = -1\).

Step 2: Calculate the discriminant

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(\Delta = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \(\Delta = (-12)^2 - 4 \cdot 9 \cdot (-1)\). Simplify the expression \(\Delta = 144 + 36 = 180\).

Step 3: Interpret the discriminant

A discriminant \(\Delta > 0\) suggests that the quadratic equation has two distinct real numbers as solutions. Since \(\Delta\) is a non-square positive number, the solutions are two irrational numbers. So, the correct choice is 'C'.

Step 4: Determine the solving method

Since the solutions are irrational numbers, it is not feasible to use the zero-factor property to solve the equation. Therefore, the quadratic formula should be used.

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations of the form ax^2 + bx + c = 0. It provides a solution to any quadratic equation, even when factoring is not possible. The quadratic formula is: x = \frac{{-b \, \pm \, \sqrt{{b^2 - 4ac}}}}{{2a}}. It allows you to solve the equation for \(x\) by simply substituting the coefficients \(a\), \(b\), and \(c\). Let's see how it's done!

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• Identify coefficients \(a\), \(b\), and \(c\) from the equation. \t
• Substitute the values into the formula: \(b\) in place of \(b\), \(a\) in place of \(a\), and \(c\) in place of \(c\). \t
• Simplify the equation under the square root first - this is called the discriminant part. \t
• Then calculate the numerator and denominator separately. \t
• Finally, solve for the two possible values of \(x\) given by the \(\pm\) sign.

. The discriminant, seen under the square root in the formula, helps us understand the nature of the solutions we will get!

zero-factor property

The zero-factor property is another important concept for solving quadratic equations, but it only works in specific cases. This property states that if the product of two expressions is zero, then at least one of the ex­pres­sions must be zero. Here's how it's used:

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• First, factorize the given quadratic equation \(ax^2 + bx + c = 0\) into a product of two binomials, if possible. \t
• Set each binomial factor equal to zero: \((x - p)(x - q) = 0\) then \(x - p = 0\) or \(x - q = 0\). \t
• Solve these simple equations to find the values of \(x\): \(x = p\) and \(x = q\).

However, in cases where factoring is difficult or when the solutions are irrational or complex numbers, this method of solving isn't applicable. This is why it's valuable to also understand how to use the quadratic formula!

solutions to quadratic equations

Solutions to quadratic equations can come in different forms based on the discriminant ( b^2 - 4ac):

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• If \(\Delta > 0\) and a perfect square, we have two distinct rational solutions. \t
• If \(\Delta > 0\) and not a perfect square, the solutions are two distinct irrational numbers. \t
• If \(\Delta = 0\), we get exactly one rational solution. \t
• If \(\Delta < 0\), the equation has two nonreal complex solutions.

Understanding the nature of the solutions helps determine the best method to solve the quadratic equation, either by factoring using the zero-factor property or by applying the quadratic formula.

irrational solutions

Irrational solutions occur when the discriminant \( b^2 - 4ac\) is positive, but not a perfect square. These solutions cannot be expressed as exact fractions or integer numbers. Instead, they remain as square roots and are considered 'irrational' because they cannot be precisely represented as a simple fraction. Here's an example to make this clearer:

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• The discriminant calculation reveals \( b^2 - 4ac = 180\) . \t
• Since 180 is positive and not a perfect square, solving using the quadratic formula \( x = \frac{{-b \, \pm \, \sqrt{{b^2 - 4ac}}}}{{2a}}\) yields solutions with square roots that cannot simplify into rational numbers. \t
• These solutions are typically left in their radical form or approximated as decimal expansions. For example, \( x = \frac{{-(-12) \, \pm \, \sqrt{{180}}}}{{2 \, 9}}\) leads us to \( x = \frac{{12 \, \pm \, \sqrt{{180}}}}{{18}}\).

Since exact fractions aren't possible, this result proves the solutions are irrational!