Question

ill give anyone a medal for helpp..

Answer 1

is there a question?

Answer 2

15. x^2+6x+1=0
16. x^2-5x-6=0
17. x^2-2x+9=0
18. 4x^2-4x=-1
19. 6x^2+5x-2=-3
20. (2x-5)^2=121

Answer 3

lets do the last one first
do you know the two square roots of \(121\) ?

Answer 4

no i do not

Answer 5

do you know a number whose square is \(121\)? it is a whole number (hint: it is not ten)

Answer 6

11?

Answer 7

yes, and also \(-11\)
therefore you have to solve two equations
\[2x-5=11\] and also solve
\[2x-5=-11\]

Answer 8

you know how to do that in two steps?

Answer 9

20)x=-3 or x=8

Answer 10

add 11 to both sides

Answer 11

no add \(5\) to both sides
\[2x-5=11\\
2x=16\\x=8\] for the first answer

Answer 12

Oh yeah

Answer 13

second one has the same steps, but a different answer
\[2x-5=-11\\
2x=-6\\
x=-3\]

Answer 14

steps were
1) add 5
2) divide by 2

Answer 15

wanna try this one? it is different
\[x^2+6x+1=0\]

Answer 16

yes

Answer 17

ok
1) subract \(1\) from both sides and get
\[x^2+6x=-1\]
then what is half of \(6\) ?

Answer 18

3?

Answer 19

yes
now we write
\[(x+3)^2=-1+3^2=-1+9=8\] or just
\[(x+3)^2=8\]

Answer 20

since you do not know what the square root of 8 is, you just write it as a radical
\[x+3=\pm\sqrt8\] don't forget the \(\pm\) cause it could be plus or minus

Answer 21

solve for \(x\) by subtracting \(3\) from both sides to finish with
\[x=-3\pm\sqrt8\]

Answer 22

if your teacher is a stickler for details, you might want to write
\[x=-3\pm2\sqrt2\] instead of \(-3\pm\sqrt8\)

Answer 23

this one \[ x^2-5x-6=0\] is easier because we can do it by factoring
do you know how to factor
\[x^2-5x-6\]?

Answer 24

16)x=-1 or x=6

Answer 25

wow what a lot of work!!
\[x^2-5x-6=(x-6)(x+1)\] and so
\[(x-6)(x+1)=0\] means either \(x-6=0\) making \(x=6\) or \(x+1=0\) giving \(x=-1\)