Question

solve quadratic equation 1 / 4 -p - 1 / 2 + p = 1 / 4​

Answer 1

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\dfrac{1}{4-p}-\dfrac{1}{2+p}=\dfrac{1}{4}$

To Find:

$\:\:\:\:\bullet\:\:\:\:\text{Value of p}$

Solution:

$:\longrightarrow\dfrac{1}{4-p}-\dfrac{1}{2+p}=\dfrac{1}{4}\\\\\text{So, taking LCM,}\\\\:\implies\dfrac{(2+p)-(4-p)}{(4-p)\times(2+p)}=\dfrac{1}{4}\\\\:\implies\dfrac{2+p-4+p}{8-2p+4p-p^2}=\dfrac{1}{4}\\\\:\implies\dfrac{2p-2}{8+2p-p^2}=\dfrac{1}{4}\\\\\text{On cross multiplying}\\\\:\implies4(2p-2)=8+2p-p^2\\\\:\implies8p-8=8+2p-p^2\\\\\text{Transposing RHS to LHS,}\\\\:\implies8p-8-8-2p+p^2=0\\\\:\implies6p-16+p^2=0\\\\:\implies p^2+6p-16=0$

$\text{\underline{METHOD1: Middle Term Splitting}}\\\\:\longrightarrow p^2+6p-16=0\\\\:\implies p^2+8p-2p-16=0\\\\:\implies p(p+8)-2(p+8)=0\\\\:\implies(p-2)(p+8)=0\\\\\bf{\underline{:\implies p=2\:\:\:\&\:\:\:-8}}$

$\text{\underline{METHOD2: Quadratic Formula}}\\\\:\longrightarrow p=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Where, a = coefficient of p ^2 , b = coefficient of p, c = constant}\\\\:\longrightarrow p^2+6p-16=0\\\\\text{Here, a = 1, b = 6, c = -16}\\\\:\implies p=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(-16)}}{2(1)}\\\\:\implies p=\dfrac{-6\pm\sqrt{36+64}}{2}\\\\:\implies p=\dfrac{-6\pm\sqrt{100}}{2}\\\\:\implies p=\dfrac{-6\pm10}{2}\\\\:\implies p=\dfrac{2\!\!\!/\:\:(-3\pm5)}{2\!\!\!/}\\\\:\implies p=-3\pm5\\\\:\implies p=-3+5\:\:\:\&\:\:\:p=-3-5\\\\\bf{\underline{:\implies p=2\:\:\:\&\:\:\:-8}}$

Verification:

$:\longrightarrow\dfrac{1}{4-p}-\dfrac{1}{2+p}=\dfrac{1}{4}\\\\\text{\underline{For p = 2,}}\\\\:\implies\dfrac{1}{4-2}-\dfrac{1}{2+2}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\\\\:\implies\dfrac{2}{4}-\dfrac{1}{4}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{4}=\dfrac{1}{4}- - - -(1) \\\\\text{\underline{For p = -8,}}\\\\:\implies\dfrac{1}{4-(-8)}-\dfrac{1}{2+(-8)}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{12}-\dfrac{1}{-6}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{12}+\dfrac{2}{12}=\dfrac{1}{4}\\\\:\implies\dfrac{3}{12}=\dfrac{1}{4}\\\\:\implies\dfrac{1}{4}=\dfrac{1}{4}- - - -(2)$

$\text{In (1) \& (2),}\\\\\text{LHS = RHS}\\\\\text{\underline{Hence verified}}$

Formula Used:

$\:\:\:\:\bullet\:\:\:\:x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Answer 2

solve quadratic equation 1 / 4 -p - 1 / 2 + p = 1 / 4

solve quadratic equation 1 / 4 -p - 1 / 2 + p = 1 / 4