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The value of $$\sqrt {32} $$ - $$\sqrt {128} $$ + $$\sqrt {50} $$ correct to 3 places of decimal is $$\sqrt {32} $$ - $$\sqrt {128} $$ + $$\sqrt {50} $$ = ?
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {\text{According to question,}} \cr
& \sqrt {32} {\text{ }} - \sqrt {128} {\text{ + }}\sqrt {50} \cr
& \Rightarrow \sqrt {16 \times 2} - \sqrt {64 \times 2} + \sqrt {25 \times 2} \cr
& \Rightarrow 4\sqrt 2 - 8\sqrt 2 + 5\sqrt 2 \cr
& \Rightarrow \sqrt 2 \cr
& \Rightarrow 1.414 \cr} $$
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