Question

answer this question...​

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:\frac{1}{4-p} - \frac{1}{2+p}=\frac{1}{4}$

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$\large\sf\underline{To\:find\::}$

• Value of p = ?

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$\large\sf\underline{Solution\::}$

$\sf\:\frac{1}{4-p} - \frac{1}{2+p}=\frac{1}{4}$

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• LCM of (4 - p) and (2 + p) = (4 - p) (2 + p)

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$\sf\implies\:\frac{[1 \times (2+p)] -[1 \times (4-p)] }{(4-p)(2+p)}=\frac{1}{4}$

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• Multiplying the terms in numerator and denominator in LHS

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$\sf\implies\:\frac{(2+p) - (4-p)}{4(2+p)-p(2+p) }=\frac{1}{4}$

$\sf\implies\:\frac{(2+p) - (4-p)}{8+4p-2p-p^{2}}=\frac{1}{4}$

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• Opening the brackets of numerator in LHS

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$\sf\implies\:\frac{2+p - 4+p}{8+4p-2p-p^{2}}=\frac{1}{4}$

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• Arranging like terms and doing simple calculations over them

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$\sf\implies\:\frac{2- 4+p+p}{8+4p-2p-p^{2}}=\frac{1}{4}$

$\sf\implies\:\frac{- 2+2p}{8+2p-p^{2}}=\frac{1}{4}$

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• Now cross multiplying

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$\sf\implies\:8+2p-p^{2}=4(-2+2p)$

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• Multiplying the terms in RHS

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$\sf\implies\:8+2p-p^{2}=-8+8p$

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• Arranging the terms in LHS

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$\sf\implies\:-p^{2}+2p+8=-8+8p$

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• Transposing + 8p to LHS it becomes - 8p

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$\sf\implies\:-p^{2}+2p-8p+8=-8$

$\sf\implies\:-p^{2}-6p+8=-8$

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• Transposing - 8 to LHS it becomes + 8

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$\sf\implies\:-p^{2}-6p+8+8=0$

$\sf\implies\:-p^{2}-6p+16=0$

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• Taking - ve sign as common

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$\sf\implies\:-(p^{2}+6p-16)=0$

$\tt\blue{\implies\:p^{2}+6p-16=0}$ ☆

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• Now we have formed a quadratic equation. Let us solve by using factorisation method

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$\sf\implies\:p^{2}+6p-16=0$

+6k in this expression can be splitted as (8p) - (2p) since :

• 6p = ( 8 - 2 ) p

• 16 = 8 × 2

This process of splitting the middle term is also known as middle term breaking .‎

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$\sf\implies\:p^{2}+(8-2)p-16=0$

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• Multiplying the terms

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$\sf\implies\:p^{2}+8p-2p-16=0$

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• Taking p as common from first two terms and -2 from second two terms

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$\sf\implies\:p(p+8)-2(p+8)=0$

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• Taking ( p + 8 ) common from whole expression

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$\sf\implies\:(p+8)(p-2)=0$

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Case 1 :

$\sf\longrightarrow\:(p+8)=\:0$

$\small{\underline{\boxed{\mathrm\red{\longrightarrow\:p=\:-8}}}}$ ★

Case 2 :

$\sf\longrightarrow\:(p-2)=\:0$

$\small{\underline{\boxed{\mathrm\red{\longrightarrow\:p=\:2}}}}$ ★

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!! Hope it helps !!