Question

Find the value of 16p^2 +1/16p^2 if 4p - 1/4p = 7

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Answer 1

❄️ Question :-

Find the value of $\sf 16p^2+\dfrac{1}{16p^2}$ if $\sf 4p-\dfrac{1}{4p}=7$.

❄️ Solution :-

Squaring on both sides :-

$\red \mapsto \sf (4p - {\Large \frac{1}{4p}})^2 =(7)^2$

Using the identity $\sf (a-b)^2=a^2+b^2-2ab$ for $\sf 4p- \dfrac{1}{4p}$

$\red \mapsto \sf {(4p)}^{2} + { \Large( {\frac{1}{4p} )}^{2}} -2(4p)({\Large \frac{1}{4p} }) = 49$

$\red \mapsto \sf 16p^2+{\Large (\frac{1}{16p^2} )}-2(\cancel{4p}) {\Large(\frac{1}{\cancel{4p}}) } = 49$

$\red \mapsto \sf (16p^2+{\Large \frac{1}{16p^2}})-2 = 49$

$\red \mapsto \sf 16p^2{\Large +\frac{1}{16p^2}}= 49 + 2$

$\red \mapsto \sf 16p^2+{\Large \frac{1}{16p^2} } = 51$

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❄️ Therefore, $\sf 16p^2+ \dfrac{1}{16p^2}=51$ if $\sf 4p-{\Large \frac{1}{4p}}=7$.

Answer 2

find the value of 16p² + 1/16p² if 4p - ¼p =7

Squaring on both sides :-

\red \mapsto \sf (4p - {\Large \frac{1}{4p}})^2 =(7)^2↦(4p−4p1)2=(7)2

Using the identity \sf (a-b)^2=a^2+b^2-2ab(a−b)2=a2+b2−2ab for \sf 4p- \dfrac{1}{4p}4p−4p1

\red \mapsto \sf {(4p)}^{2} + { \Large( {\frac{1}{4p} )}^{2}} -2(4p)({\Large \frac{1}{4p} }) = 49↦(4p)2+(4p1)2−2(4p)(4p1)=49

\red \mapsto \sf 16p^2+{\Large (\frac{1}{16p^2} )}-2(\cancel{4p}) {\Large(\frac{1}{\cancel{4p}}) } = 49↦16p2+(16p21)−2(4p)(4p1)=49

\red \mapsto \sf (16p^2+{\Large \frac{1}{16p^2}})-2 = 49↦(16p2+16p21)−2=49

\red \mapsto \sf 16p^2{\Large +\frac{1}{16p^2}}= 49 + 2↦16p2+16p21=49+2

\red \mapsto \sf 16p^2+{\Large \frac{1}{16p^2} } = 51↦16p2+16p21=51