Question

If the roots of the quadratic equation $x^2 - 2px + q^2 = 0$ are real and distinct, then:

• A. $|p| > |q|$
• B. $|p| < |q|$
• C. $p^2 \ge q^2$
• D. $p^2 \le q^2$

Hint:

"Real and distinct" strictly translates to a strict inequality ($>$). This immediately rules out options (C) and (D) containing $\ge$ and $\le$.

Answer 1

The Correct Option is A

Step 1: The rule for two different real roots.
A quadratic has two real and distinct roots only when its discriminant $D = b^2 - 4ac$ is strictly positive.

Step 2: Read off the coefficients.
For $x^2 - 2px + q^2 = 0$ we have $a = 1$, $b = -2p$, and $c = q^2$.

Step 3: Write the discriminant.
Plug in: \[ D = (-2p)^2 - 4(1)(q^2) = 4p^2 - 4q^2 \]

Step 4: Apply the condition.
For distinct real roots we need $D > 0$: \[ 4p^2 - 4q^2 > 0 \]

Step 5: Simplify.
Divide by 4: \[ p^2 > q^2 \]

Step 6: Use the size form.
$p^2 > q^2$ is the same as comparing absolute values: \[ |p| > |q| \] \[ \boxed{ |p| > |q| } \]