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

 

❓ 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 Solution

We have a quadratic:

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

Compare with $x^2 - Sx + P = 0$, where

• Sum of roots $= S = p+2$
  

• Product of roots $= P = 2p+9$
  

For both roots to be negative real numbers, we need:

1. Real roots → Discriminant $\mathrm{\Delta }\ge 0$

2. Both roots negative →

   • Sum of roots $<0$
     

   • Product of roots $>0$
     

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🔹 Step 1 — Conditions from sum and product

Sum < 0:

$$p+2<0\Rightarrow p<-2$$

Product > 0:

$$2p+9>0\Rightarrow p>-\frac{9}{\mathrm{2}}$$

Together:

$$-\frac{9}{2}<p<-2$$

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🔹 Step 2 — Discriminant condition

$$\mathrm{\Delta }=(p+2{)}^2-4(2p+9)\ge 0$$

Compute:

$$\mathrm{\Delta }=(p^2+4p+4)-(8p+36)=p^2-4p-32$$

So we solve:

$$p^2-4p-32\ge 0$$

Roots of $p^2 - 4p - 32 = 0$:

$$p=\frac{4\pm \sqrt{16+128}}{2}=\frac{4\pm \sqrt{144}}{2}=\frac{4\pm 12}{\mathrm{2}}$$

So,

$$p=8\text{or}p=-4$$

Quadratic opens upwards, so

$$p\le -4\text{or}p\ge 8$$

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🔹 Step 3 — Combine all conditions

We already have from sum & product:

$$-\frac{9}{2}<p<-2$$

Intersect this with:

$$p\le -4\text{or}p\ge 8$$

Only the overlap is:

$$\boxed{{\displaystyle -\frac{9}{2}<p\le -\mathrm{4}}}$$

So the interval is:

$$(\alpha ,\beta ]=(-\frac{9}{2},-4]$$

Thus:

$$\alpha =-\frac{9}{2},\beta =-4$$

Now compute:

$$\beta -2\alpha =-4-2(-\frac{9}{2})=-4+9=5$$

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🧮 Image Solution

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✅ Conclusion & Video Tip

✅ Final Answer:

$$\boxed{5}$$