Question

Determine the value(s) of k such that each trinomial is a perfect square.
a) x^2 + 4x + k
b) 4x^2 - 12x + k

Answer 1

two steps only

Answer 2

what is half of 4?

Answer 3

@misty1212 half of 4 is 2

Answer 4

what's next?

Answer 5

\[b^2-4ac=0\]

Answer 6

\[4^2-4(1)(k)=0\]

Answer 7

k=?

Answer 8

12?

Answer 9

Try again

Answer 10

4^2-4(1)(k)
16-4k=0
divide both sides by -4
k=-4

Answer 11

is that right?

Answer 12

\[16-4k=0 \\ \\ 16=4k \\ \\ k=?\]

Answer 13

k = 4
oops

Answer 14

thank you

Answer 15

now how about part b

Answer 16

Do the same for the next one
\[4x^2 - 12x + k=0 \\ Using \\ ax^2+bx+c=0 \\ \\ b^2-4ac=0\]

Answer 17

\[(-12)^2-4(4)(k)=0\]

Answer 18

-144-16k=0
-144=16k
divide both sides by 16
k=-9 ?

Answer 19

\[(-12)^2= (-12)(-12)=12 \times 12=144\]

Answer 20

k=9

Answer 21

oh i see it's because i didn't put (-12)^2 in brackets

Answer 22

thanks for the help though!