rationalising the denominator 7/3-√2
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Answered by
1
Step-by-step explanation:
The given equation is:
\frac{7-3\sqrt{2}}{7+3\sqrt{2}}
7+3
2
7−3
2
Rationalizing the denominator, we have
\frac{7-3\sqrt{2}}{7+3\sqrt{2}}{\times}\frac{7-3\sqrt{2}}{7-3\sqrt{2}}
7+3
2
7−3
2
×
7−3
2
7−3
2
\frac{(7-3\sqrt{2})^{2}}{(7)^{2}-(3\sqrt{2})^2}
(7)
2
−(3
2
)
2
(7−3
2
)
2
\frac{(7-3\sqrt{2})^{2}}{49-18}
49−18
(7−3
2
)
2
\frac{(7-3\sqrt{2})^{2}}{31}
31
(7−3
2
)
2
which is the required rationalized form.
Answered by
0
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