Find Parameter Range for Negative Roots

 

❓ Question

Let the set of all values of $p \in \mathbb{R}$ for which both the roots of the equation

$$x^2-(p+2)x+(2p+9)=0$$

are negative real numbers be the interval $(\alpha, \beta]$.
Then the value of

$$\beta -2\alpha$$

is equal to ?

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Quadratic equations
👉 Nature of roots
👉 Conditions for negative roots.

Main idea:

For both roots to be negative real numbers:

Conditions:

$$\boxed{ D\ge0 }$$
$$\boxed{ \alpha+\beta<0 }$$
$$\boxed{ \alpha\beta>0 }$$

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🔷 Step 1 — Compare with Standard Form 💯

Given:

$$x^2-(p+2)x+(2p+9)=0$$

For quadratic:

$$x^2-Sx+P=0$$

we have:

$$\text{Sum of roots}=S=p+2$$
$$\text{Product of roots}=P=2p+9$$

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🔷 Step 2 — Condition for Negative Roots

Both roots negative:

Sum must be negative

$$p+2<0$$
$$\boxed{ p<-2 }$$

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Product must be positive

$$2p+9>0$$
$$\boxed{ p>-\frac92 }$$

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🔷 Step 3 — Real Roots Condition

Discriminant:

$$D=(p+2)^2-4(2p+9)$$
$$=p^2+4p+4-8p-36$$
$$=p^2-4p-32$$
$$=(p-8)(p+4)$$

For real roots:

$$D\ge0$$

Thus:

$$p\le-4 \quad\text{or}\quad p\ge8$$

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🔷 Step 4 — Combine All Conditions

We need simultaneously:

$$p<-2$$
$$p>-\frac92$$
$$p\le-4 \text{ or } p\ge8$$

Only common interval:

$$\boxed{ \left(-\frac92,-4\right] }$$

Thus:

$$\alpha=-\frac92$$
$$\beta=-4$$

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🔷 Step 5 — Calculate Required Value

$$\beta-2\alpha$$
$$=-4-2\left(-\frac92\right)$$
$$=-4+9$$
$$\boxed{ 5 }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Only discriminant use kar lena

❌ Negative roots ke conditions bhool jaana

Remember:

For both roots negative:

$$\boxed{ \text{Sum}<0,\quad \text{Product}>0 }$$

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✅ Final Answer

$$\boxed{ 5 }$$

(Option 4)​

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