Question

16x^2-24x-13 =0 solve it by quadric
method

Answer 1

Given:

p(x) = 16x²–24x–13 = 0

✪✪Solution✪✪

According to the quadratic formula:

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

Therefore, accordingly:

$⇒x = \frac{ - ( - 24)± \sqrt{( - 24) ^{2} - 4(16)( - 13) } }{2(16)}$

$⇒x = \frac{24± \sqrt{576 + 832} }{32}$

$⇒x = \frac{24± \sqrt{1408} }{32}$

$⇒x = \frac{24±8 \sqrt{22} }{32}$

$⇒x = \frac{3± \sqrt{22} }{4}$

Thus, the two possible value of 'x' are:

$⇒x = \frac{3 - \sqrt{22} }{4} \: \: \: and \: \: \: \frac{3 + \sqrt{22} }{4}$

Answer 2

Answer: 16x^2 -24x-13

It has no real zeros.

Explanation: