Question

if A and B are the zeros of the quadratic polynomial FX equal to x square - 4 x + Q then find the value of a 4+ b4

Answer 1

$f(x) = {x}^{2} - 4x + q$
So as we know its roots are a and b hence
$a + b \: = \frac{ - coeffeicient \: of \: x}{coefficient \: of \: {x}^{2} }$
hence a + b = -(-4)/1 = 4
then
$ab = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$
ab = Q/1 = Q
Now
${a}^{2} + {b}^{2} = {(a + b)}^{2} \: - 2ab$
Now input the values
${a}^{2} + {b}^{2} = {4}^{2} - 2(q) \\ {a}^{2} + {b}^{2} = 16 - 2q \\ \\ {a}^{4} + {b}^{4} = ({a}^{2} + {b}^{2} ) {}^{2} - 2 {(ab)}^{2} \\ {a}^{4} + {b}^{4} = (16 - 2q ) {}^{2} - 2 {q}^{2} \\ {a}^{4} + {b}^{4} = 256 + 4 {q}^{2} - 32q - 2 {q}^{2} \\ {a}^{4} + {b}^{4} \: = 256 + 2 {q}^{2} - 32q$
If you are given the value of Q then you can input it.
Hope this helps you and if it possible then please mark it as the brainliest answer.