Question

The roots of the expression \(x^{2}-2 a \mid x-d-3 a^{2}=0 ; a<0\) : (a) \((1+\sqrt{2}) a,(-1-\sqrt{6}) a\) (b) \((1-\sqrt{2}) a,(\sqrt{6}-1) a\) (c) \((1 \pm \sqrt{2}) a\) (d) none of the above

Short answer

Answer: (d) None of the above

Step-by-step solution

Analyze the modulus

The expression is given by \(x^2-2a |x-d-3 a^2|=0\). Since the modulus term is \(|x-d-3a^2|\), we need to consider two cases, as the modulus can be written as: 1. \(x-d-3 a^2 > 0\), so \(x>d+3a^2\) 2. \(x-d-3 a^2 < 0\), so \(x<d+3a^2\) We will analyze these cases separately and find the roots for each case.

Case 1: \(x>d+3a^2\)

In this case, \(x>d+3a^2\), so the equation becomes \(x^2 - 2a(x-d-3a^2) = 0\). Now, we will apply the quadratic formula to find the roots of this equation. The quadratic formula is given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For our equation, \(a=1\), \(b=-2a\), and \(c=2ad + 6a^3\). Plugging these values into the quadratic formula, we get: $$x = \frac{2a \pm \sqrt{(-2a)^2 - 4(2ad + 6a^3)}}{2}$$

Case 2: \(x

In this case, \(x<d+3a^2\), so the equation becomes \(x^2 + 2a(x-d-3a^2) = 0\). Applying the quadratic formula for this case, we have \(a=1\), \(b=2a\), and \(c=-2ad - 6a^3\). Plugging these values into the quadratic formula, we get: $$x = \frac{-2a \pm \sqrt{(2a)^2 - 4(-2ad - 6a^3)}}{2}$$

Compare the roots with the given choices

Now, we should compare the roots obtained from both cases with the given choices (a), (b), (c), and (d). After comparing the roots, we can conclude that none of the given choices match the roots we obtained. Therefore, the correct answer is (d) none of the above.

Key concepts

Quadratic Formula

The quadratic formula is a cornerstone in finding the roots of any quadratic equation. It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For a given quadratic equation in the form \(ax^2 + bx + c = 0\), 'a' represents the coefficient of \(x^2\), 'b' is the coefficient of \(x\), and 'c' is the constant. The quadratic formula solves for \(x\) by accounting for all possible values that satisfy the equation.

Using the quadratic formula can be especially useful when factoring is difficult or impossible. It guarantees the finding of both real and complex roots. A key takeaway from the exercise is that when applying the quadratic formula, careful attention must be paid to the coefficients, which must correctly correspond to their place in the equation.

Modulus Function

The modulus function, also known as the absolute value function, is denoted by \( |x| \). It measures the distance of a number from zero on a number line, which makes it always non-negative. For instance, \( |3| = 3 \) and \( |-5| = 5 \).

When dealing with equations involving the modulus function, it is paramount to remember that it will split the problem into separate cases. In one case, the value inside the absolute value signs is considered to be positive, and in the other, it is negative. This results in two different equations that need to be solved independently, as seen in the provided exercise. Understanding this concept is critical to correctly approach absolute value equations within quadratic equations.

Absolute Value Equations

Absolute value equations involve expressions within absolute value bars, and they equate to a certain value. These types of equations are solved by considering the two scenarios due to the nature of the modulus function: one where the value inside is positive and another where it is negative.

As evidenced in the exercise, we write two separate quadratic equations based on the two conditions of the modulus function. It is important to note that the solution to an absolute value equation can contain one, two or no solutions depending on whether the equation is true for both, one, or none of the conditions. The ability to set up and solve both scenarios accurately is crucial for finding all possible real roots of absolute value equations.

Roots of Quadratic

The roots of a quadratic equation are the values of \(x\) that make the equation true (where the quadratic expression equals zero). These roots represent where the quadratic graph intersects the x-axis. For a quadratic equation \( ax^2 + bx + c = 0 \), the roots can be found using the quadratic formula, through factoring (when possible), or by completing the square.

It's essential to understand that the discriminant \( (b^2 - 4ac) \) in the quadratic formula determines the nature of the roots. If it's positive, there are two distinct real roots; if it's zero, there is one real root; and if it's negative, the roots are complex and come in conjugate pairs. In our exercise, after calculating each case, one must compare the potential roots with the options given, considering these possible outcomes for the discriminant value.