Question

4a^2-16a+17=0 solve this pls​

Answer 1

$4 {x}^{2} - 16x + 17 = 0 \\ x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{16 \frac{ + }{} \sqrt{256 - 272} }{2} = \frac{16 \frac{ + }{} \sqrt{ - 16} }{8} \\ x = \frac{16 \frac{ + }{ }4 i}{8} = \frac{4 \frac{ + }{}i }{2}$

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Answer 2

---------- QUADRATIC FORMULA FOR $ax^2+2b'x+c=0$ ----------

$2b'$ stands for the even number coefficient.

Apply quadratic formula first.

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

$=\frac{-2b'\pm 2\sqrt{b'^2-ac} }{2a}$

$=\frac{-b'\pm \sqrt{b'2-ac} }{a}$

∴ $x=\frac{-b'\pm \sqrt{b'^2-ac} }{a}$ are the roots of a quadratic equation $ax^2+2b'x+c=0$.

------------------------------ Using the formula ------------------------------

$x=\frac{8\pm \sqrt{64-68} }{4} =\frac{8\pm 2i }{4} =\frac{4\pm i}{2}$