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Maths help needed asap

how would you solve this???
y=5-x^2-4/x^2
please someone help me
any help much appreciated
Reply 1
Avatar for Kshm2686
Kshm2686
OP
11
that’s the original question:
C78BB5C2-66EE-4BC5-AE44-BC599024D236.jpg.jpeg
Reply 2
Avatar for happily
happily
10
do you remember integration.
to find the area (actual value), you need to identify a and b first, which are two x values that encloses the area.
in the question, a and b values are two x values when y=0.
Reply 3
Avatar for sotor
sotor
14
id use the table function on my calculator to figure out where f(x) equals zero and then integrate with those limits

i honestly wouldn't know how to solve the equation as is!
Reply 4
Avatar for Kshm2686
Kshm2686
OP
11
Original post by happily
do you remember integration.
to find the area (actual value), you need to identify a and b first, which are two x values that encloses the area.
in the question, a and b values are two x values when y=0.


integration part is easy but I’m just confused about f(x) how would u solve it when u make y=0? what are the x values?
Reply 5
Avatar for Kshm2686
Kshm2686
OP
11
Original post by sotor
id use the table function on my calculator to figure out where f(x) equals zero and then integrate with those limits

i honestly wouldn't know how to solve the equation as is!


Thank you!
Reply 6
Avatar for happily
happily
10
Original post by mantekes
integration part is easy but I’m just confused about f(x) how would u solve it when u make y=0? what are the x values?


so, when you rewrite the equations in linear form (no fraction) -> y=5 - x^2 - 4x^-2
to find the roots, when y=0, x^2 -5 + 4^x-2 = 0
because of the awkward negative power,
multiply them by x^2 so it becomes:
x^4 - 5x^2 + 4 = 0
then, you use substitution-ish method (forgot the name), for example, let a^2 = x^4, which means:
x^4 - 5x^2 + 4 = 0 becomes a^2 - 5a + 4 = 0
Reply 7
Avatar for Kshm2686
Kshm2686
OP
11
Original post by happily
so, when you rewrite the equations in linear form (no fraction) -> y=5 - x^2 - 4x^-2
to find the roots, when y=0, x^2 -5 + 4^x-2 = 0
because of the awkward negative power,
multiply them by x^2 so it becomes:
x^4 - 5x^2 + 4 = 0
then, you use substitution-ish method (forgot the name), for example, let a^2 = x^4, which means:
x^4 - 5x^2 + 4 = 0 becomes a^2 - 5a + 4 = 0


Thank you sooo much!! it makes so much more sense now! thank you!
Reply 8
Avatar for happily
happily
10
Original post by mantekes
Thank you sooo much!! it makes so much more sense now! thank you!


no problem! :smile:
after finding roots of the coded equation, eg a=1, a=4, dont forget that a^2=x^4 so a=x^2
which means x = root(1) and root(4)

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