HEY FRIENDS ... PLEASE HELP
PLEASE FRIENDS...
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For 3 to be the root of the given equation, we must make LHS and RHS equal to zero.
Let's see how.....
LHS =>>>
√(x^2-4x+3) + √(x^2-9) = √(9-12+9) + √(9-9)
= 0.
RHS==>>>
√(4x^2 -14x + 16) = √(36-52+16)
= 0.
Hence, LHS = RHS = 0.
Therefore, 3 is the root/ zero of the given equation.
Hope it helps❤️❤️
Let's see how.....
LHS =>>>
√(x^2-4x+3) + √(x^2-9) = √(9-12+9) + √(9-9)
= 0.
RHS==>>>
√(4x^2 -14x + 16) = √(36-52+16)
= 0.
Hence, LHS = RHS = 0.
Therefore, 3 is the root/ zero of the given equation.
Hope it helps❤️❤️
Answered by
0
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