Question

Solve and show steps : q + 12 – 2(q – 22) > 0

Answer 1

q + 12 - 2(q - 22) > 0
q + 12 - 2q - 44 > 0
q -32 - 2q > 0
-32 minus \[2q^{2}\] >0
subract -32 from 0 or add 32 to 0 same thing
\[-2q^{2}\] > 32
then square root the 2 from both sides
\[\sqrt{-2q}\]
> \[\sqrt{32}\]
(sorry i cant figure out how to get some of the equations on the same line)
once that is done then it will leave u with -2q< \[\sqrt{32}\]
you can solve it further but i was tought to leave it it like that if the square root doesnt come out a whole number

Answer 2

Ok , thanks. :)

Answer 3

q + 12 - 2q +44 > 0 Distributed the -2
56 - q > 0 grouped like terms
q < 56

Answer 4

@kal94 is incorrect

Answer 5

you should not have a squared term

Answer 6

also they did not distribute correctly

Answer 7

oh ok .

Answer 8

i was using a calculator and thats what it gave me, ill check to see if i typed in a wrong number or symbol

Answer 9

q + 12 - 2q +44 > 0 [Distributed the -2]
56 - q > 0 [grouped like terms]
q < 56 [added q to each side]

To check you can always plug in a number greater than 56 and it shouldn't work and a number less than and it should work.

Answer 10

@kal94 just look at the equation. the only way to get a q^2 is when there is q*q. This doesn't exist above. Something is wrong

Answer 11

i realized that now i pressed multiplication symbol instead of plus on the calculator

Answer 12

That would do it