Quadratic Equations
Equations involving modulus
Grade 11
Question:
<p>86. If the equation \(x^2 + ax + b = 0\) has distinct real roots and \(x^2 + a|x| + b = 0\) has only one real root, then which of the following is true?</p>
<p>(1) \(b = 0, a > 0\)</p>
<p>(2) \(b = 0, a < 0\)</p>
<p>(3) \(b > 0, a < 0\)</p>
<p>(4) \(b < 0, a > 0\)</p>
Step-by-Step Solution
Key Concept: If x²+a|x|+b=0 has only one real root, then by substitution y=|x|, we get y²+ay+b=0 must have exactly one positive root (y>0), since each positive y gives two values of x=±y. The original equation having distinct real roots constrains the coefficients.
<p><strong>Step 1:</strong> For x²+ax+b=0 to have distinct real roots: a²-4b>0, so a²>4b</p><p><strong>Step 2:</strong> For x²+a|x|+b=0, substitute y=|x|≥0 to get y²+ay+b=0. This equation has only one real root in x iff y²+ay+b=0 has exactly one non-negative root.</p><p><strong>Step 3:</strong> For one non-negative root in y: either (i) one root is 0 and other is negative, or (ii) both roots exist with product≤0. This requires b·(coefficient)≤0, so b≤0.</p><p><strong>Step 4:</strong> Also, if b<0, the roots of y²+ay+b=0 have opposite signs (product is negative). The positive root y₀>0 gives x=±y₀ (two roots in x), plus we need exactly one real root total in the original equation in x. This happens when b=0 and a<0, OR when discriminant condition is: <strong>b<0 and a>0</strong>.</p><p><strong>Step 5:</strong> Combining: a²>4b and b<0 with a>0 gives us <strong>b<0</strong> and <strong>a>0</strong>.</p><p>∴ Answer: b < 0 (or equivalently a > 0 and b < 0, or ab < 0)</p>
Correct Answer: 2