Question

find the zero of f(x)=4x^2-17​

Answer 1

$f(x) = 4 {x}^{2} - 17$

Since a zero of polynomial f(x), is the value of x for which f(x) is zero

Therby,

$4 {x}^{2} - 17 = 0 \\ by \: applying \: quadratic \: formula \\ x = \frac{ - b \: + - \: \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{+ - \sqrt{ - 4(4)( - 17)} }{8} \\ x = + - \frac{ 4 \sqrt{17} }{8} \\ x = + \frac{ \sqrt{17} }{2} \\ x = - \frac{ \sqrt{17} }{2}$

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