Problem 8 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If the roots of the equation \(x^{2}-8 x+\left(a^{2}-6 a\right)\) \(=0\) are real, then (a) \(-2

Step-by-Step Solution

Answer

The correct choice is (c) \(-2 \leq a \leq 8\).

Step 1: Understanding the Condition for Real Roots

For the quadratic equation \( ax^2 + bx + c = 0 \), the roots are real if the discriminant \( b^2 - 4ac \geq 0 \). For our given equation \( x^2 - 8x + (a^2 - 6a) = 0 \), we have \( a = 1 \), \( b = -8 \), and \( c = a^2 - 6a \). Thus, the discriminant is: \((-8)^2 - 4(1)(a^2 - 6a) \geq 0\).

Step 2: Simplifying the Discriminant Expression

Calculate the discriminant expression: \((-8)^2 - 4 \cdot 1 \cdot (a^2 - 6a)\). This simplifies to: \[ 64 - 4(a^2 - 6a) \]. Expand the expression: \[ 64 - 4a^2 + 24a \]. This results in the inequality: \[-4a^2 + 24a + 64 \geq 0\].

Step 3: Rearranging and Solving the Quadratic Inequality

Rearrange the inequality \(-4a^2 + 24a + 64 \geq 0\) by dividing through by \(-4\) (which reverses the inequality sign): \[ a^2 - 6a - 16 \leq 0 \]. Factor the quadratic: \( (a - 8)(a + 2) \leq 0 \).

Step 4: Finding Interval for \(a\) Using the Critical Points

Determine the critical points by setting each factor equal to zero: \( a - 8 = 0 \implies a = 8 \) and \( a + 2 = 0 \implies a = -2 \). Test intervals around these critical points to see where the expression \((a - 8)(a + 2) \leq 0\) holds. The solution to the inequality is: \( -2 \leq a \leq 8 \).

Step 5: Choose the Correct Option

Compare the solution interval \( -2 \leq a \leq 8 \) to the given options. The correct choice is (c) \(-2 \leq a \leq 8\).

Key Concepts

Discriminant

The discriminant is a key part of the quadratic formula, and it tells us how many and what kind of roots a quadratic equation has. The discriminant \(\Delta\) in the quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(b^2 - 4ac\). This simple expression helps in determining whether the solutions (roots) of the equation are real or complex.

If the discriminant is:

• Greater than zero, the quadratic equation has two distinct real roots.
• Equal to zero, the quadratic equation has exactly one real root, which is also called a repeated or double root.
• Less than zero, the quadratic equation has no real roots, instead it has two complex roots.

In our specific problem, the equation \(x^2 - 8x + (a^2 - 6a) = 0\) involved calculating the discriminant as part of determining when the roots are real. We found that as long as the discriminant \(64 - 4(a^2 - 6a) \geq 0\), the roots are real.

Real Roots

Real roots are the solutions of a quadratic equation that are real numbers, as opposed to imaginary or complex numbers. When we solve a quadratic equation graphically, real roots are the x-intercepts of the quadratic function's graph, which is a parabola.

For an equation \(ax^2 + bx + c = 0\), the roots can be determined by solving the quadratic formula: \((-b \pm \sqrt{b^2 - 4ac})/(2a)\). The discriminant, \(b^2 - 4ac\), dictates the nature of these roots. This exercise required us to confirm the condition under which the discriminant provided real roots for a specific quadratic equation. By solving the inequality \(-4a^2 + 24a + 64 \geq 0\), we ensured the roots were real, leading to the interval \(-2 \leq a \leq 8\).

Quadratic Inequalities

Quadratic inequalities involve quadratic expressions that are either greater than or equal to, or less than or equal to zero. These inequalities are crucial for determining the range of values that satisfy a particular quadratic condition, like having real roots.

In our context, we derived the inequality \(-4a^2 + 24a + 64 \geq 0\) from the discriminant condition for real roots. Solving this inequality involved rearranging it to \(a^2 - 6a - 16 \leq 0\) by dividing by \(-4\) and remembering to flip the inequality sign.

To solve the quadratic inequality, we factor it to get \((a - 8)(a + 2) \leq 0\). Critical points \(a = 8\) and \(a = -2\) mark the boundaries where the expression changes its sign. Testing intervals around these critical points helped us determine where the expression remains non-positive, leading to the solution set \(-2 \leq a \leq 8\).

Factoring Quadratics

Factoring is a technique used to simplify quadratic expressions or solve quadratic equations. It involves expressing the quadratic in terms of its roots. A factored quadratic looks like \( (px + q)(rx + s) = 0\), where each factor represents a potential solution to the equation.

In our problem, once we rearranged the quadratic inequality to \(a^2 - 6a - 16 \leq 0\), the next step was to factor it. The factors were found to be \(a - 8\) and \(a + 2\), meaning the equation can be expressed as \((a - 8)(a + 2) = 0\). These factors give the critical points that are used to solve the inequality.

Knowing how to factor quadratics is essential because it simplifies the process of finding roots and solving inequalities. Factoring converts a complex quadratic into a product of simpler binomials, making it easier to analyze and solve for the variable.