Let the set of all values of p \in \mathbbR , for which both the roots of the equation x^2-(p+2) x+(2 p+9)=0 are negative real numbers, be the interval (\alpha, \beta] . Then \beta-2 \alpha is equal to

Quadratic Equation and Inequalities PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q37 with Solution

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Correct answer: A

We need both roots to be real and negative.

For a quadratic $x^2-Sx+P=0$ , if roots are $r_1,r_2$ , then:

\qquad r_1r_2=P.$$ Here, $$r_1+r_2=p+2, \qquad r_1r_2=2p+9.$$ Also, for real roots, discriminant must satisfy $$D\ge 0.$$ --- 2. Condition for real roots: $$D=(p+2)^2-4(2p+9)\ge 0.$$ Simplify: $$p^2+4p+4-8p-36\ge 0$$ $$p^2-4p-32\ge 0$$ $$(p-8)(p+4)\ge 0.$$ Hence, $$p\le -4 \quad \text{or} \quad p\ge 8.$$ --- 3. Condition for both roots to be negative. If both roots are negative, then: - sum of roots must be negative: $$p+2<0 \implies p<-2,$$ - product of roots must be positive: $$2p+9>0 \implies p>-\frac{9}{2}.$$ So from these two, $$-\frac{9}{2}<p<-2.$$ --- 4. Combine with the discriminant condition. We need $$p\in \left(-\frac{9}{2},-2\right)$$ and also $$p\le -4 \quad \text{or} \quad p\ge 8.$$ Intersection gives: $$p\in \left(-\frac{9}{2},-4\right].$$ Why is $p=-4$ included? At $p=-4$, discriminant is zero, so both roots are equal. The equation becomes $$x^2-(-2)x+(1)=x^2+2x+1=(x+1)^2,$$ so the repeated root is $-1$, which is negative. Hence $p=-4$ is allowed. Thus, $$(\alpha,\beta]=\left(-\frac{9}{2},-4\right].$$ So, $$\alpha=-\frac{9}{2},\qquad \beta=-4.$$ --- 5. Compute: $$\beta-2\alpha=-4-2\left(-\frac{9}{2}\right)=-4+9=5.$$ Therefore, the required value is $$\boxed{5}.$$

View written solutionFree

Correct answer: A

We need both roots to be real and negative.

For a quadratic $x^2-Sx+P=0$ , if roots are $r_1,r_2$ , then:

\qquad r_1r_2=P.$$ Here, $$r_1+r_2=p+2, \qquad r_1r_2=2p+9.$$ Also, for real roots, discriminant must satisfy $$D\ge 0.$$ --- 2. Condition for real roots: $$D=(p+2)^2-4(2p+9)\ge 0.$$ Simplify: $$p^2+4p+4-8p-36\ge 0$$ $$p^2-4p-32\ge 0$$ $$(p-8)(p+4)\ge 0.$$ Hence, $$p\le -4 \quad \text{or} \quad p\ge 8.$$ --- 3. Condition for both roots to be negative. If both roots are negative, then: - sum of roots must be negative: $$p+2<0 \implies p<-2,$$ - product of roots must be positive: $$2p+9>0 \implies p>-\frac{9}{2}.$$ So from these two, $$-\frac{9}{2}<p<-2.$$ --- 4. Combine with the discriminant condition. We need $$p\in \left(-\frac{9}{2},-2\right)$$ and also $$p\le -4 \quad \text{or} \quad p\ge 8.$$ Intersection gives: $$p\in \left(-\frac{9}{2},-4\right].$$ Why is $p=-4$ included? At $p=-4$, discriminant is zero, so both roots are equal. The equation becomes $$x^2-(-2)x+(1)=x^2+2x+1=(x+1)^2,$$ so the repeated root is $-1$, which is negative. Hence $p=-4$ is allowed. Thus, $$(\alpha,\beta]=\left(-\frac{9}{2},-4\right].$$ So, $$\alpha=-\frac{9}{2},\qquad \beta=-4.$$ --- 5. Compute: $$\beta-2\alpha=-4-2\left(-\frac{9}{2}\right)=-4+9=5.$$ Therefore, the required value is $$\boxed{5}.$$