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Showing posts with the label both roots negative condition

JEE Main Shortcut: Kab Quadratic ke Dono Roots Negative Hote Hain? 💡

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  ❓ Concept Both Roots Negative Trick in 59 Sec Finding when both roots of a quadratic equation are negative is one of the most scoring and repetitive JEE topics . Good news? 👉 It follows just 3 golden rules . 1️⃣ Quadratic Form General quadratic: a x 2 + b x + c = 0 ax^2 + bx + c = 0 First, ensure: a > 0 a > 0 📌 If a < 0 a < 0 , multiply the equation by –1 before applying the rules. 2️⃣ Condition 1 — Roots Must Be Real For real roots: D = b 2 − 4 a c ≥ 0 D = b^2 - 4ac \ge 0 This condition is mandatory . No real roots → no sign discussion. 3️⃣ Condition 2 — Sum of Roots α + β = − b a \alpha + \beta = -\frac{b}{a} For both roots negative : α + β < 0 \alpha + \beta < 0 Since a > 0 a > 0 : − b a < 0 ⇒ b > 0 -\frac{b}{a} < 0 \Rightarrow b > 0 4️⃣ Condition 3 — Product of Roots α β = c a \alpha\beta = \frac{c}{a} For both roots negative : α β > 0 \alpha\beta > 0 Again, since a > 0 a > 0 : c > 0 c > 0 5️⃣...

Roots Negative? Here’s the Fastest Way to Solve This p-Parameter Question ⚡

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  ❓ Question Let the set of all values of p ∈ R p \in \mathbb{R}  for which both the roots of the equation x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 are negative real numbers be the interval ( α , β ] (\alpha, \beta] . Then the value of β − 2 α is equal to ? 🖼️ Question Image ✍️ Short Solution We have a quadratic: x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 Compare with x 2 − S x + P = 0 x^2 - Sx + P = 0 , where Sum of roots = S = p + 2 = S = p+2 Product of roots = P = 2 p + 9 = P = 2p+9 For both roots to be negative real numbers , we need: Real roots → Discriminant Δ ≥ 0 Both roots negative → Sum of roots < 0 <0 Product of roots > 0 >0 🔹 Step 1 — Conditions from sum and product Sum < 0: p + 2 < 0 ⇒ p < − 2 Product > 0: 2 p + 9 > 0 ⇒ p > − 9 2​ Together: − 9 2 < p < − 2 🔹 Step 2 — Discriminant condition Δ = ( p + 2 ) 2 − 4 ( 2 p + 9 ) ≥ 0 Compute: Δ = ( p 2 + 4 p + 4 ) − ( 8 p + 36 ) = p 2 − 4 p − 32 So...