Question

The least integral value of 'a' for which the
equation x2
$x {}^{2}$
– 2(a – 1) x +(2a + 1) = 0 has
both the roots positive is
(A) 3
(B) 4
(C) 1
(D) 5​

Answer 1

Solution

Given: An equation $x^2-2(a-1)x+2a+1=0$.

To find: The least integral value of $a$ that makes both the root positive.

For roots to be a real number,

$\dfrac{D}{4} =(a-1)^2-(2a+1)=a^2-2a+1-2a-1$

$=a^2-4a=\boxed{a(a-4)\geq 0}$

Let the roots be $\alpha ,\beta$.

First, the sum and product of two positive numbers are positive.

$\begin{cases} & \alpha +\beta =2a-2 \\ & \alpha \beta =2a+1 \end{cases}$

We need both of them to be positive.

$\implies\begin{cases} & a>1 \\ & a>-\dfrac{1}{2} \end{cases}$

$\implies \boxed{a>1}$

Now we have two inequality, and both need to be satisfied, so

$\begin{cases} & a\leq 0, a\geq 4 \\ & a> 1 \end{cases}\implies \boxed{a\geq 4}$

The least integral value in this inequality is 4. So the correct option is (B).

Answer 2

Given :-

$\sf x^2-2(a-1)x+(2a+1)$

To Find :-

Intergal value

Solution :-

Since both roots are roots are positive

$\sf f(0) > 0$

By putting value

$\sf f(0) = (0)^2 - 2(a-1)0+(2a+1)>0$

$\sf f(0) = 0 - 0 + 2a+1>0$

$\sf f(0) = 2a+1>0$

$\sf f(0)=\dfrac{1}{2}+a>0$

$\sf f(0) = a>0-\dfrac{1}{2}$

$\sf f(0) = a>\dfrac{-1}{2}$

$\sf f(0) = 2a>-1$

$\sf f(0)=\dfrac{2(a-1)}{2}>0$

$\sf f(0)=a>1$

$\sf 4(a-1)^2-4(2a+1)\geq0$

$\sf (a-1)^2-2a-1\geq0$

$\sf a^2-2a+1-2a-1\geq0$

$\sf a^2-2a-2a\geq0$

$\sf a^2-4a\geq0$

$\sf a(a-4)\geq0$

$\sf a =4$