a^1/2 +b^1/2 -c^1/2 =0, then the value of (a+b-c)^2
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Answered by
13
Given, √a+√b -√c = 0
⇒ √a+√b = √c Squaring on both sides
⇒ (√a+√b)² = √c²
⇒ a+b+2√ab = c
⇒ (a+b-c) = -2√ab
Again squaring on both sides, we get
(a+b-c)² = (-2√ab)²
∴ (a+b-c)² = 4ab
⇒ √a+√b = √c Squaring on both sides
⇒ (√a+√b)² = √c²
⇒ a+b+2√ab = c
⇒ (a+b-c) = -2√ab
Again squaring on both sides, we get
(a+b-c)² = (-2√ab)²
∴ (a+b-c)² = 4ab
Answered by
2
Answer:
Step-by-step explanation:
√a +√b -√c=0
√a+√b= √c
Squaring both sides
(√a+√b)²= √c²
a+b+2√ab =c
a+b-c = -2√ab
Again on squaring
(a+b-c)² = (-2√ab)²
(a+b-c)= 4ab
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