If the quadratic equation ( +2)x^2-3 x+4 =0, -2, has two positive roots, then the number o…

Question

If the quadratic equation $(\lambda +2){x}^{2}-3\lambda x+4\lambda =0,\lambda 
e -2$, has two positive roots, then the number of possible integral values of $\lambda $ is :

Accepted Answer

For the quadratic equation

$(\lambda +2){x}^{2}-3\lambda x+4\lambda =0,\, \lambda 
e -2$

to have two positive roots, let us use the NCERT conditions for a quadratic equation.

If the roots are $\alpha $ and $\beta $, then for both roots to be positive, we need:

Discriminant $\ge 0$ so that roots are real,

Sum of roots $>0$,

Product of roots $>0$.

1. Identify coefficients
Comparing with $a{x}^{2}+bx+c=0$, we get

$a=\lambda +2,\, b=-3\lambda ,\, c=4\lambda $

2. Condition for real roots
Discriminant:

$D={b}^{2}-4ac$

So,

$D=(-3\lambda {)}^{2}-4(\lambda +2)(4\lambda )$

$D=9{\lambda }^{2}-16\lambda (\lambda +2)$

$D=9{\lambda }^{2}-16{\lambda }^{2}-32\lambda $

$D=-7{\lambda }^{2}-32\lambda $

For real roots,

$-7{\lambda }^{2}-32\lambda \ge 0$

$-\lambda (7\lambda +32)\ge 0$

Multiplying by $-1$ reverses the inequality:

$\lambda (7\lambda +32)\le 0$

This gives

$-\frac{32}{7}\le \lambda \le 0$

So possible integral values from this condition are

$\lambda =-4,-3,-2,-1,0$

But given $\lambda 
e -2$, so remaining are

$\lambda =-4,-3,-1,0$

3. Condition for sum of roots positive
Sum of roots is

$\alpha +\beta =-\frac{b}{a}=\frac{3\lambda }{\lambda +2}$

For two positive roots,

$\frac{3\lambda }{\lambda +2}>0$

Since $3>0$, this means

$\frac{\lambda }{\lambda +2}>0$

This is positive when numerator and denominator have same sign:

$\lambda >0$ and $\lambda +2>0$, or

$\lambda

So,

$\lambda >0\, \, \lambda From our shortlisted values $-4,-3,-1,0$, this keeps only

$\lambda =-4,-3$

4. Condition for product of roots positive
Product of roots is

$\alpha \beta =\frac{c}{a}=\frac{4\lambda }{\lambda +2}$

For positive roots,

$\frac{4\lambda }{\lambda +2}>0$

Again,

$\frac{\lambda }{\lambda +2}>0$

This gives the same condition:

$\lambda >0\, \, \lambda Among the shortlisted values, again we get

$\lambda =-4,-3$

5. Check both values

For $\lambda =-4$:

$D=-7(16)-32(-4)=-112+128=16>0$

So roots are real and distinct.

For $\lambda =-3$:

$D=-7(9)-32(-3)=-63+96=33>0$

So roots are real and distinct.

Hence both values give two positive roots.

Therefore, the number of possible integral values of $\lambda $ is

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