The least integral value of ‘a’ for which the
equation x2
[tex]x {}^{2} [/tex]
– 2(a – 1) x +(2a + 1) = 0 has

July 15, 2021

By Isabelle

2 thoughts on “The least integral value of ‘a’ for which the<br />equation x2<br />[tex]x {}^{2} [/tex]<br /> – 2(a – 1) x +(2a + 1) = 0 has<br /”

1. Given :-

   [tex]\sf x^2-2(a-1)x+(2a+1)[/tex]

   To Find :-

   Intergal value

   Solution :-

   Since both roots are roots are positive

   [tex]\sf f(0) > 0[/tex]

   By putting value

   [tex]\sf f(0) = (0)^2 – 2(a-1)0+(2a+1)>0[/tex]

   [tex]\sf f(0) = 0 – 0 + 2a+1>0[/tex]

   [tex]\sf f(0) = 2a+1>0[/tex]

   [tex]\sf f(0)=\dfrac{1}{2}+a>0[/tex]

   [tex]\sf f(0) = a>0-\dfrac{1}{2}[/tex]

   [tex]\sf f(0) = a>\dfrac{-1}{2}[/tex]

   [tex]\sf f(0) = 2a>-1[/tex]

   [tex]\sf f(0)=\dfrac{2(a-1)}{2}>0[/tex]

   [tex]\sf f(0)=a>1[/tex]

   [tex]\sf 4(a-1)^2-4(2a+1)\geq0[/tex]

   [tex]\sf (a-1)^2-2a-1\geq0[/tex]

   [tex]\sf a^2-2a+1-2a-1\geq0[/tex]

   [tex]\sf a^2-2a-2a\geq0[/tex]

   [tex]\sf a^2-4a\geq0[/tex]

   [tex]\sf a(a-4)\geq0[/tex]

   [tex]\sf a =4[/tex]

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2. Solution

   Given: An equation [tex]x^2-2(a-1)x+2a+1=0[/tex].

   To find: The least integral value of [tex]a[/tex] that makes both the root positive.

   For roots to be a real number,

   [tex]\dfrac{D}{4} =(a-1)^2-(2a+1)=a^2-2a+1-2a-1[/tex]

   [tex]=a^2-4a=\boxed{a(a-4)\geq 0}[/tex]

   Let the roots be [tex]\alpha ,\beta[/tex].

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

   [tex]\begin{cases} & \alpha +\beta =2a-2 \\ & \alpha \beta =2a+1 \end{cases}[/tex]

   We need both of them to be positive.

   [tex]\implies\begin{cases} & a>1 \\ & a>-\dfrac{1}{2} \end{cases}[/tex]

   [tex]\implies \boxed{a>1}[/tex]

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

   [tex]\begin{cases} & a\leq 0, a\geq 4 \\ & a> 1 \end{cases}\implies \boxed{a\geq 4}[/tex]

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

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