Question

Let 'k' and 's' be the zeroes of the polynomial P(x) = x^2 +2ax + a-1. Then the least value of (k-s)^2 is ?

Anyone here can solve this problem?

Answer 1

$p(x) = {x}^{2} + 2ax + (a - 1) \\ k + s = \frac{ - 2a}{1} \\ ks = a - 1 \\ (k - s) {}^{2} = (k + s) {}^{2} -4 ks \\ (k - s) {}^{2} = 4 {a}^{2} - 4(a - 1) \\ (k - s) {}^{2} = 4( {a}^{2} - a + 1) \\ k - s = 2 \sqrt{ {a}^{2} - a + 1 }$
for (k-s) to be minimum... the right hand side time should have to be minimum....

rhs contains square root,, and square root can give minimum zero... I. e., non negative value....

but the function I. e., quadratic inside square root has minimum value of (3/4) ....i. e, at a=1/2

therefore minimum value of square root =√3/2

$min(k - s) = mi \: n \: (2 \sqrt{ {a}^{2} - a + 1)} \\ k - s = 2 \times (√3/2) \\ k - s = √3$

MARK AS BRAINLIEST IF HELPED

for calculating minimum value of a^2-a+1
f(a)=a^2-a+1
complete the square
f(a)=(a-(1/2))^2-(1/4)+1
minimum value of square =0
min f(a)=1-(1/4)
f(a)=3/4

and =√3