If the equation $x^8 – 3abx + a^2 = 0$ has one root that is $\ge 3$. And if $a\geq 0$ .
Can $b$ be equal to 15?
I took One root of this equation is (3 + t), where t ≥ 0 And put the value of this root in equation.
Thanks in advance.
If the equation $x^8 – 3abx + a^2 = 0$ has one root that is $\ge 3$. And if $a\geq 0$ .
Can $b$ be equal to 15?
I took One root of this equation is (3 + t), where t ≥ 0 And put the value of this root in equation.
Thanks in advance.
Hint: Try to solve for $a$ in terms of $x$ when $b=15$
Your solution should include a square root. Can you spot anything about this square root when $x \ge 3$?