Question

how many real solutions does x have if 81x^2 + 64< 144x ?

Answer 1

Step-by-step explanation:

We have,

$81 {x}^{2} + 64 < 144x$

$\implies81 {x}^{2} - 144x + 64 < 0$

$\implies {(9x)}^{2} - 2 \times 9x \times 8 + ( {8})^{2} < 0$

$\implies(9x - 8)^{2} < 0$

This isn't possible, because square of any number is always positive,

So, no real value of x satisfies the given inequality