Question

(2a-3)^+(3b-5)^2+(4c-7)^2 = 0, then find the value of sqrt(4a+3b+4c)

Answer 1

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Answer 2

❤❤Here is your answer ✌ ✌

$\huge\underline {\red {\bold {Answer}}}$

(2a-3)^+(3b-5)^2+(4c-7)^2 = 0

(2a-3)^2=0
2a-3=0
a=3/2

(3b-5)^2=0
3b=5
b=5/3

(4c-7)^2 = 0
4c=7
c=7/4

now

(4a+3b+4c)
4×3/2+3×5/3+4×7/4
root 18
=>>3 root 2