Question

88. Solve the
following quadratic equation
by using quadratic formula..
cx²+ (c+d) x + d=0, c not equal to 0​

Answer 1

ANSWER:

To Solve:

cx² + (c + d)x + d = 0

Solution:

$\text{We are given that,}\\\\:\longrightarrow cx^2+(c+d)x+d=0\\\\\text{We know that,}\\\\\text{For a quadratic equation, Ax^2+Bx+C=0 ,}\\\\\text{By Quadratic Formula,}\\\\:\hookrightarrow x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}\\\\\text{So, in this case, A = c, B = (c + d), C = d}\\\\\text{Hence,}\\\\:\implies x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}\\\\:\implies x=\dfrac{-(c+d)\pm\sqrt{(c+d)^2-4(c)(d)}}{2(c)}\\\\\text{We know that,}\\\\:\hookrightarrow (p+q)^2=p^2+q^2+2pq\\\\\text{So,}$

$:\implies x=\dfrac{-(c+d)\pm\sqrt{c^2+d^2+2cd-4cd}}{2c}\\\\:\implies x=\dfrac{-(c+d)\pm\sqrt{c^2+d^2-2cd}}{2c}\\\\\text{We know that,}\\\\:\hookrightarrow p^2+q^2-2pq=(p-q)^2\\\\\text{So,} \\\\:\implies x=\dfrac{-c-d\pm\sqrt{(c-d)^2}}{2c}\\\\:\implies x=\dfrac{-c-d\pm(c-d)}{2c}\\\\:\implies x=\dfrac{-c\!\!\!/-d+c\!\!\!/-d}{2c}\:\:or\:\:x=\dfrac{-c-d\!\!\!/-c+d\!\!\!/}{2c}\\\\:\implies x=\dfrac{-2\!\!\!/\:d}{2\!\!\!/\:c}\:\:or\:\:x=\dfrac{-2c\!\!\!\!\!/}{2c\!\!\!\!\!/}\\\\:\implies x=\dfrac{-d}{c}\:\:and\:\:x=-1\\\\\text{So,}\\\\\bf{:\implies x=\dfrac{-d}{c}\:\:\:and\:\:\:-1}$

Formula Used:

$\text{1)For a quadratic equation, ax^2+bx+c=0 ,}\\\\\text{By Quadratic Formula,}\\\\:\hookrightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\:\hookrightarrow2)\: (p\pm q)^2=p^2+q^2\pm2pq$