Question

5s^2-42s+49=0

Answer 1

Pr 245 sum -42
5s^2-35s-7s+49=0
5s(s-7)-7(s-7)=0
(5s-7)(s-7)=0
5s-7=0 or s-7=0
s=7/5 or s=7

Answer 2

got it thanks

Answer 3

Check: 5(7/5)^2-42*7/5+49=0
5*49/25-294/5+49=0
49/5-294/5+49=0
-49+49=0
And do the same for the second answer s=7 to make sure it's correct.

Answer 4

how about 6x^2-19+10=0

Answer 5

You are welcome.
Try to follow the same steps. Do you understand what to do?

Answer 6

i tryied but i could seem to factor it right

Answer 7

OO. this one little different.

Answer 8

so it is 15 and 4

Answer 9

there is a x behind 19

Answer 10

6x^2-19+10=0
(6x -4 )(6x -15 )/6
=2(6x - 4)3(2x - 15)/6
=(6x-4)(2x-15)

Answer 11

6x^2-9=0 factor out 3
3(2x^2-3)=0
2x^2-3=0
2x^2=3
X^2=3/2
X=+-sqrt(3/2)

Answer 12

it was solution for 19.

Answer 13

sorry i wrote problem wrong is is 6x^2-19x+10=0

Answer 14

I figured that.

Answer 15

the problem really started as 6x^2=19x-10

Answer 16

Yes, I get it. so try same steps again. Product is 60, sum is -19

Answer 17

so would be -15 and -4 right

Answer 18

yes, and then we will rewrite -19x = -15x -4x

Answer 19

then you will factor by grouping and have the result.

Answer 20

You always can check your answers by plugging them into the original equation and make sure it's becomes true statements.

Answer 21

thanks

Answer 22

okay how about x^2+10ax+21a^2

Answer 23

i don't even know where to start

Answer 24

(x + 3a )(x + 7a)

Answer 25

try factoring without the "a" first, then figure out what a should be to make the equation make sense visually.

Answer 26

sweet thanks

Answer 27

You are welcome.
For the SECOND problem answers are (3x-2)(2x-5)
x=2/3 or x=5/2
Sorry need to go.

Answer 28

how about 3x^2-10xy+8y^+8y^2=0

Answer 29

wrote it wrong 3x^2-10xy+8y^2

Answer 30

(3x - 6y)(3x -4y )/3
3(x - 2y)(3x - 4y)/3
(x - 2y)(3x - 4y)

Answer 31

more please :)

Answer 32

2(s-4)^2=128

Answer 33

(s-4)^2 - 64 = 0
( (s-4) - 8)( (s -4 ) + 8) = 0
(s - 12)(s + 4) = 0

Ans. s = 12 and s = -4