Question

Use the quadratic formula to find the roots of the equation. Round to tenths if necessary. x2 - 6x + 2 = 0
A. {0.4, 5.6}
B. {0.5, 4.8}
C. {1.4, 2.4}
D. no real number solution

Answer 1

That isn't a perfect square.

Answer 2

A
do you want me to post the explanation ?

Answer 3

yes......

Answer 4

\[\Large \begin{array}{l}
\frac{\begin{array}{l}
{x^2} - 6x + 2 = 0\\
- b \pm \sqrt {{b^2} - 4ac}
\end{array}}{{2a}}\\
\frac{{6 \pm \sqrt {{{( - 6)}^2} - 4(1 \cdot 2)} }}{{2 \cdot 1}}\\
= \frac{{6 \pm \sqrt {28} }}{2}\\
= {\rm{5}}{\rm{.65,0}}{\rm{.35}}
\end{array}\]

Answer 5

fewscrewsmissing did it ;)

Answer 6

I don't understand how you get 2 values......

Answer 7

That's the roots of the equation - plug either into the polynomial, and it'll equate to 0. (NB: I've rounded here, so THOSE particular numbers won't be 0; but the actual explicit solution).

Answer 8

pratu043 notice the +/-

Answer 9

ok thanks

Answer 10

plus or minus

Answer 11

pratu043 the "quadratic formula" is well worth remembering - precisely for when factoring is difficult.