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19 solutions
really Llewellyn you made me understood well and i am acknowledging you so much
Using the identity, ( a 2 − b 2 ) = ( a + b ) ( a − b ) , the fraction can be easily factorized and solved.
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But how do you go from (11²-9²)=20x2? Shouldn't it be (11²-9²) = (121-81) = 40? I know they're the same answer but where did that 20x2 come from?
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It's factorised. (a2-b2) = (a+b)(a-b) = (20)(2)
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Yes, using that factorization many times :) That's what I liked about this problem.
(Of course, one could expand out everything, but that's not the point.)
here again they have used the formula (a^2 - b^2) = (a+b)(a-b) i.e (11+9)(11-9) = 20x2 =40
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He got (11+9) which is equal to 20 and then he got (11-9) which is equal to 2 so he got (20)(2) which is just 20*2 which is equal to 40.
Yes, essentially they are just factoring one too many times to make it confusing. After you are at 11^2 - 9^2, you could just as easily say 11^2 = 121 and 9^2 = 81 so 121-81 = 40. They were trying to get you to use the "rules" for factoring/ FOIL etc. when in this specific case it would still be practical to just solve as I showed above.
From (11^2-9^2)==(11+9)×(11-9)=20×2=40 Now u get it??
a^2-b^2 = (a+b)(a-b). So (11^2-9^2) = (11+9)(11-9), which also equals (20)(2) because of PEMDAS: basically do the operations within the parentheses, and finally multiply.
Quite easy x^4-y^4/x^2+y^2 = (x^2-y^2)(x^2+y^2)/(x^2+x^2)= x^2-y^2 Put values and get the ans
yes, and we could use it once more, to make it easier to calculate :)
11^4-9^4=
(11)^2-(9)^2
=(11^2+9^2)(11^2-9^2) [a^2-b^2=(a+b)(a-b)]
so,(11^2+9^2)(11^2-9^2)/(11^2+9^2)
=11^2-9^2
=121-81
40
(11^4-9^4)/11^2+9^2=(11^2-9^2) (11^2+9^2)/(11^2+9^2)=(11^2-9^2) =(11+9) (11-9)=20*2=40
I thought it is 11^4-9^4=2^4 and 11^2+9^2=20^2 and simplify 2^4/20^2, simplify again 2^2/20 =4/20 or 1/5.. my answer is 1/5, andi think all answers are wrong, this is only my idea, correct me if im wrong... thanks
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One easy way to show that it isn't correct is to use smaller values e.g. 3^4 - 2^4. From your solution, the answer would be 3^4 - 2^4 = 1^4=1. However, 3^4 = 81 and 2^4 = 16. Therefore 3^4 - 2^4 = 81 - 16 = 65. You can now see that your method wouldn't work. Also, it can be simplified as 3^4 - 2^4 = (3^2 - 2^2)(3^2 + 2^2) = (9 - 4)(9 + 4) = 5 x 13 = 65. Re-solve the question using this method
No, look up Rules of Exponents . You are saying that a n − b n = ( a − b ) n and a n + b n = ( a + b ) n , neither of which is true.
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I think that you are right.. The answers above, of the exercise are wrong.. It should be 4 the final result.
a² - b²= (a+b)(a-b) So . 11⁴ — 9⁴ ÷ 11²+9²= (11²—9²)(11²+9²)÷(11²+9²) = 11²—9² =121—81 =40
1 1 2 + 9 2 1 1 4 − 9 4 = 1 1 2 + 9 2 ( 1 1 2 − 9 2 ) ( 1 1 2 + 9 2 ) = 1 1 2 − 9 2 = 1 2 1 − 8 1 = 4 0
11^4-9^4/11^2+9^2 = (14641-6561)/(121+81) = 8080/202=40
I know this isn't the prettiest way to do it at, but brute force works most of the times. ;)
11^4 / 11^2 = 11^2
-9^4 / 9^2 = -9^2
11^2 - 9^2 = 121 - 81 = 40
cut the same terms, the remaining is 11^2-9^2 which is equal to 40 And that's Answer.....thanks!
(11^4 - 9^4)/(11^2 + 9^2) = {(11^2 - 9^2). (11^2 + 9^2)}/(11^2 - 9^2) = 11^2 + 9^2 = 121 - 81 = 40
(11² + 9²)x(11² - 9²)/(11² + 9²) = 11² - 9² = (11+9)(11-9) = 20x2 = 40
Using the difference of two squares identity, we know that 1 1 4 − 9 4 = ( 1 1 2 + 9 2 ) ( 1 1 2 − 9 2 ) . Because the ( 1 1 2 + 9 2 ) cancel, we are left with ( 1 1 2 − 9 2 ) , or 40.
That's nice :)
[(11^2+9^2)(11^2-9^2)]/(11^2+9^2) =(11^2-9^2) =40
That's nice :)
We knw dat a^2 -b^2= (a+b)(a-b).. use the identity and solve it...
Yes indeed. Can you add more details?
(11^4-9^4)/(11^2+9^2)=11^2-9^2 =(11+9)(11-9)=20×2=40
11^2^2-9^2^2/11^2+9^2 =(11^2+9^2)(11^2-9^2)/11^2+9^2) =11^2-9^2 =121-81 =40