Given
x 4 - 3x 2 + k =0
we want to find value of k, for which the equation, x 4 - 3x 2 + k =0 has two distinct roots in the interval [2,3].
we write equation in this form
( x 2 ) 2 − 3 ( x 2 ) + k = 0 (x^2)^2-3(x^2)+k=0 ( x 2 ) 2 − 3 ( x 2 ) + k = 0
let two distinct root is x1 and x2
according to question
2 ≤ x 1 , x 2 ≤ 3 4 ≤ x 1 2 , x 2 2 ≤ 9 2\leq x_1, x_2 \leq 3\\
4\leq x_1^2, x_2^2 \leq 9\\ 2 ≤ x 1 , x 2 ≤ 3 4 ≤ x 1 2 , x 2 2 ≤ 9
we know ,
x = − b ± b 2 − 4 a c 2 a x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\\ x = 2 a − b ± b 2 − 4 a c
hence
4 ≤ 3 ± 9 − 4 k 2 ≤ 9 8 ≤ 3 ± 9 − 4 k ≤ 18 5 ≤ ± 9 − 4 k ≤ 15....... ( 1 ) n o w t a k e + o r − o n e b y o n e 5 ≤ 9 − 4 k ≤ 15.... ( 2 ) a n d 5 ≤ − 9 − 4 k ≤ 15 o r − 5 ≥ 9 − 4 k ≥ − 15..... ( 3 ) f r o m ( 2 ) a n d ( 3 ) t h e r e n o e x i s t i n g k w h i c h s a t i s f i e d e q ( 2 ) a n d ( 3 ) 4\leq \frac{3\pm\sqrt{9-4k}}{2} \leq 9\\
8\leq {3\pm\sqrt{9-4k}}{} \leq 18\\
5\leq \pm\sqrt{9-4k}\leq 15.......(1)
\\
now \space take \space + or- \space one by one\\
5\leq \sqrt{9-4k}\leq 15....(2)\\
and\\
5\leq -\sqrt{9-4k}\leq 15\\
or\\
-5\geq \sqrt{9-4k}\geq -15.....(3)\\
from(2) and(3)\\
there\space no\space existing\space k \space which \space \\satisfied \space eq(2)and(3)
\\ 4 ≤ 2 3 ± 9 − 4 k ≤ 9 8 ≤ 3 ± 9 − 4 k ≤ 18 5 ≤ ± 9 − 4 k ≤ 15....... ( 1 ) n o w t ak e + or − o n e b yo n e 5 ≤ 9 − 4 k ≤ 15.... ( 2 ) an d 5 ≤ − 9 − 4 k ≤ 15 or − 5 ≥ 9 − 4 k ≥ − 15..... ( 3 ) f ro m ( 2 ) an d ( 3 ) t h ere n o e x i s t in g k w hi c h s a t i s f i e d e q ( 2 ) an d ( 3 )
hence we say that
there is no real number, k for which the equation, x 4 - 3x 2 + k =0 has two distinct roots in the interval [2,3].
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