Question

Math Help 5 questions Multiple Choice

Answer 1

@amistre64 @happinessbreaksbones

Answer 2

i believe all of these can be appraoched pretty simply using the quadratic formula

Answer 3

given: \(ax^2+bx+c=0\)

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 4

Doing part 1 by completing the square, \[(x+3)^2-2=0 --> (x+3)^2=2 --> x+3 = ±\sqrt{2} --> x= -3±\sqrt{2}\]

Answer 5

you can line feed in latex code by typing \\

Answer 6

and \to is a simple \rightarrow
\[(x+3)^2-2=0\\ \to (x+3)^2=2 \\\to x+3 = ±\sqrt{2}\\ \to x= -3±\sqrt{2}\]

Answer 7

@TripleC can you help me with the rest?

Answer 8

Sorry about this taking too long, but just tried to do it completing the square and it was nearly 50 lines, just gonna use quadratic forumla and talk you through that instead xD

Answer 9

k :)

Answer 10

is 4 d? and is 3 d and 2 i don't know

Answer 11

As amistre64 said the quadratic forumla is \[x= \frac{ -b \pm \sqrt{b ^{2-4ac}} }{ 2a }\]
a is the coefficent (number before) x^2 so in this case a=2
b is the coeffienct of x b=-8
c is the constant which is 7

You want all the terms on the same side to use this forumla so that's why c is 7 not -7 as it was moved.

So we just substitute all the values in to get \[\frac{ 8 \pm \sqrt{64-4x2x7}}{ 4 }\]

Which can can simplfy to \[\frac{8\pm \sqrt{8} }{ 4 }\]

\[\sqrt{8}\] can be simplfied to \[\sqrt{4}* \sqrt{2} = 2*\sqrt{2} = 2\sqrt{2}\]

So we now have \[\frac{ 8\pm 2\sqrt{2} }{ 4 } = \frac{ 8 }{ 4 } \pm \frac{ 2\sqrt{2} }{ 4 } = 2\pm\frac{ \sqrt{2} }{ 2 }\]

Answer 12

is 2 d as well

Answer 13

i knew it oky is 3 correct?

Answer 14

i got d aswell

Answer 15

Are you saying that is the 4th option for question 3 correct?

Answer 16

for#4 i have d for #3 i also have d am i correct?

Answer 17

@TripleC can you just do #4 step by step please

Answer 18

What do you mean by d :s? And by 4 you mean the one on the right of 2?