If -4 is a root of 2x^+px- 12=0. Find value of p.
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Answered by
0
is it x^2??
if it is so then
(-4)^2+4p-12=0
16-4p-12=0
4-4p=0
4p= 4
p=1
and if it is 2×x^2
Then, 2×(-4)^2-4p-12=0
32-4p-12=0
4p=20
So, p=5
Plz mark it as brainliest answer
if it is so then
(-4)^2+4p-12=0
16-4p-12=0
4-4p=0
4p= 4
p=1
and if it is 2×x^2
Then, 2×(-4)^2-4p-12=0
32-4p-12=0
4p=20
So, p=5
Plz mark it as brainliest answer
Answered by
2
2*(-4)^2+p(-4)-12=0
2*16-4p=12
32-4p=12
-4p=12-32
-4p=-20
p=5
2*16-4p=12
32-4p=12
-4p=12-32
-4p=-20
p=5
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