Question

what are the x intercepts of the quadratic equation 2x^2-8x+3

Answer 1

do you know the quadratic formula?

Answer 2

no i forgot it

Answer 3

\(\Large x\; =\; \frac {-b\; \pm\; \sqrt {b^2\; -\; 4ac}}{2a} \)
a=2 b=-8 c=3

Answer 4

and this will help me find both x intercepts ?

Answer 5

yes!
to find solutions/zeros/roots/x-int for a quadratic equation: \(a x^2\; +\; b x\; +\; c\; =\; 0 \), we always use the quadratic formula.

Answer 6

im stuck i at this section i have: \[-(-8)\pm \sqrt{\frac{ 40 }{ 4 }}\] what do i do after this ?

Answer 7

\(\Large x\; =\; \frac {-(-8)\; \pm\; \sqrt {(-8)^2\; -\; 4(2)(3)}}{2(2)} \\\Large x\; =\; \frac {-(-8)\; \pm\; \sqrt {64\; -\;24}}{4}\\\Large x\; =\; \frac {8\; \pm\; \sqrt {40}}{4}\)
break it down
\(\Large x\; =\; \frac {8\; +\; \sqrt {40}}{4}\) and \(\Large x\; =\; \frac {8\; -\; \sqrt {40}}{4}\)
then solve it, you've got two x-int (ofc since it is a quadratic function)

Answer 8

since 40 doesn't have a square root how do i solve it ?

Answer 9

i usually don't ask for the answer but i didn't go to the last class where we discussed this so i am completely lost so could you help me out and just explain your answer so i understand how you got there

Answer 10

\(\sqrt{40}=2 \sqrt{10}=6.32\)
\(\Large {x\; =\; \frac {8\; +\; \sqrt {40}}{4}\ and \ \Large x\; =\; \frac {8\; -\; \sqrt {40}}{4}\\x= \; \frac {4\; +\; \sqrt {10}}{2}=3.58 \ or \ x= \; \frac {4\; -\; \sqrt {10}}{2}=0.42}\)
there you go , done :)
it's 3.58 or 0.42

Answer 11

you can use either the decimal or the simplified form of the radical term

Answer 12

do you have any confusions about this topic?

Answer 13

no you were a big help thank you so much i really appreciate the help

Answer 14

can i ask you one more question ?

Answer 15

sorry i was afk,
okay sure, i'll try :)