Question

let a>2 be a constant if there are just 18 positive integers satisfying the inequality (x-a)(x-2a)(x-a in to the pawer 2)<0 then the value of a? - Google Search

Answer 1

$(x - a)(x - 2a) {(x - a)}^{2} \: < \: 0$

${(x - a)}^{2} will \: always \: be \: greater \: than \: zero \:$

So,

$(x - a)(x - 2a) < \: 0$

Now, if (x-a) will be positive then (x-2a) should be negative to satisfy the above inequality.

$x > a \: \: and \: \: x < 2a$

And if (x-a) will be negative then (x-2a) should be positive to satisfy the above inequality.

$x < a \: \: and \: \: x > 2a$

Now, from a common solution of the above conditions we get,

a = 19

So,

$(x - 19)(x - 38) {(x - 19)}^{2} \: < \ 0$
So total of 18 positive integers are the solution of above inequality,

x = {20,21,22,23,.....36,37}