One of my CAT exam problem simplifies:
9x^4 - 4y^4 = (3x^2)^2 - (2y^2)^2.
Can someone explain this rule for me? I simplified 9x^4 to be (3^2)x^4...and that got me nowwhere. But I'm trying to memorize the correct rule.
rockrock Wrote:One of my CAT exam problem simplifies:
9x^4 - 4y^4 = (3x^2)^2 - (2y^2)^2.
Can someone explain this rule for me? I simplified 9x^4 to be (3^2)x^4...and that got me nowwhere. But I'm trying to memorize the correct rule.
julien.pitteloud Wrote:Hello Aditya,
could we continue the rule for
(3x^2-2y^2)=(sqrt(3)x-sqrt(2)y)(sqrt(3)x+sqrt(2)y)
but do you know if this is sometimes used to iterate further (exponents 1/2, 1/4, aso) ?? This would have no sense.
rockrock Wrote:If 9x^4 ---> 3x^2 * 3x^2 (multiply bases and add the exponents)
rockrock Wrote:Ok Let me rephrase this a bit...
The example I used was 2^2 * 2^2 = (2^2)^2 which is 2^4. Here, the same base/add exponent rule applies, correct?
But with 3x^2 * 3x^2 -- the result is 9x^4, not 3x^4...., we dont retain the 3x base as I did in the above example. Is it because there are two integers 3 and x - that we don't retain the base?
Or is the rule that if you have same base/same exponent you are supposed to square them and not apply the add exponent rules.
so
step 1: 3^2 * 3^2 =
step 2: (3^2)^2 =
step 3: 3^4.
You cant skip the 2nd step when another variable (x) is involved.