Question

find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me how this is done.

Answer 1

For a quadratic of the form $f(x)=ax^2+bx+c$, we have the quadratic formula 
$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given $2x^2-16x+27=0$, we can get that $a=2, \ b=-16$, and $c=27$. We substitute these numbers into the quadratic formula above.

$x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}$

$x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}$

$x=\dfrac{16 \pm \sqrt{40} }{4}$

$x=\dfrac{16 \pm 2\sqrt{10} }{4}$

$x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}$

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the $\pm$ symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.