The Student Room Group

Coordinate Systems

Question:

Workings:

(33p2)2+36p2 \sqrt{(3-3p^2)^2 + 36p^2}
=9p4+18p2+9 = \sqrt{9p^4 + 18p^2 + 9}
=3p2+32p+3 = 3p^2 + 3\sqrt{2}p + 3

That is what I have ended up with. Why is this incorrect?
(edited 7 years ago)
Original post by NinCheng
Question:

Workings:

(33p2)2+36p2 \sqrt{(3-3p^2)^2 + 36p^2}
=9p4+18p2+9 = \sqrt{9p^4 + 18p^2 + 9}
=3p2+32p+3 = 3p^2 + 3\sqrt{2}p + 3

That is what I have ended up with. Why is this incorrect?



You cannot do that with square roots, a+ba+b\sqrt{a + b} \neq \sqrt{a} + \sqrt {b} in general.

Instead, try factorising what you have under the square root sign.
Reply 2
Original post by NinCheng
Question:

Workings:

(33p2)2+36p2 \sqrt{(3-3p^2)^2 + 36p^2}
=9p4+18p2+9 = \sqrt{9p^4 + 18p^2 + 9}
=3p2+32p+3 = 3p^2 + 3\sqrt{2}p + 3

That is what I have ended up with. Why is this incorrect?


As above. Do you think that 2=1+1=1+1=1+1=2\sqrt{2} = \sqrt{1 + 1} = \sqrt{1 } + \sqrt{1} = 1 + 1 = 2? Surely not.

Instead, 9p4+18p2+9=3p4+2p2+1=3(p2)2+2(p2)+1\sqrt{9p^4 + 18p^2 + 9} = 3\sqrt{p^4 + 2p^2 + 1} = 3\sqrt{(p^2)^2 + 2(p^2) + 1}
Reply 3
Original post by 16Characters....
You cannot do that with square roots, a+ba+b\sqrt{a + b} \neq \sqrt{a} + \sqrt {b} in general.

Instead, try factorising what you have under the square root sign.


Which leaves me with 9(p4+2p2+1) \sqrt{9(p^4 + 2p^2 + 1)} but now I'm clueless
Original post by NinCheng
Which leaves me with 9(p4+2p2+1) \sqrt{9(p^4 + 2p^2 + 1)} but now I'm clueless


Can you see how to factorise that bracket? You might like to let y = p^2
Reply 5
line 2, take a common factor of 9 out of expression which becomes 3 outside sqrt. The function left inside the sqrt is a perfect square.

Your fundamental error is that a+b\sqrt{a+b} does NOT equal a+b\sqrt a + \sqrt b which is what your "solution' assumes.

if
a+b\sqrt{a+b} did equal a+b\sqrt a + \sqrt b then try it with a and b = 1 which would predict that the square root of 2 was 2
Reply 6
Original post by 16Characters....
Can you see how to factorise that bracket? You might like to let y = p^2


Oh, so

9(p4+2p2+1) \sqrt{9(p^4 + 2p^2 + 1)}
9(p2+1)(p2+1) \sqrt{9(p^2 + 1)(p^2 + 1)}
9(p2+1)2 \sqrt {9(p^2+1)^2}
=3(p2+1) = 3(p^2 + 1)
Original post by NinCheng
Oh, so

9(p4+2p2+1) \sqrt{9(p^4 + 2p^2 + 1)}
9(p2+1)(p2+1) \sqrt{9(p^2 + 1)(p^2 + 1)}
9(p2+1)2 \sqrt {9(p^2+1)^2}
=3(p2+1) = 3(p^2 + 1)


Yep.
Reply 8
Original post by 16Characters....
Yep.


Thank you :smile:

Quick Reply

Latest