Radical expressions or absolute value?
1 0 − 4 6 can also be represented as __________ .
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2 solutions
The problem can be rephrased as
(
6
)
2
+
(
4
)
2
−
6
4
which is in the form of
a
2
+
b
2
−
2
a
b
⇒
(
a
+
b
)
2
⇒
a
+
b
So the answer is
6
−
2
.
Careful, ( a + b ) 2 = ∣ a + b ∣ which does NOT necessarily equal a + b .
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but square root of a negative number is a complex number and square root of a positive number(or zero) is always positive or zero.When you do ( a + b ) 2 it always leads to a positive number or zero.So there is no question of taking the absolute value of a + b .
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I'm not saying that the radicand is negative, I'm saying that a radical can only give a positive value. What if a + b is negative? You still have a positive radicand but you must take the positive version of a + b once you take the radical of ( a + b ) 2
And you made typo in answer.
1 0 − 4 6 = 1 0 − 2 2 4 . We know that ( a − b ) 2 = a + b − 2 a b . Comparing with original equation, a + b = 1 0 and a b = 2 4 . Clearly a = 6 and b = 4 satisfies this. So its equivalent to ( 6 − 2 ) 2 . Its PRINCIPAL square root is equal to 6 − 2