The value of x such that $4 (1 + y) x^{2} - 4 x y + 1 - y = 0$ $4(1+y)x^2-4xy+1-y=0$ is?

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Answer 1 · 2018-07-23T06:36:26

Considering the given eqution with a change

$4 (1 + y) x^{2} - 4 x y - (1 - y)$ $4(1+y)x^2-4xy-(1-y)$

$\Rightarrow 4 (1 + y) x^{2} - 2 (1 + y) x + 2 (1 - y) x - (1 - y)$ $=>4(1+y)x^2-2(1+y)x+2(1-y)x-(1-y)$

$\Rightarrow 2 (1 + y) x (2 x - 1) + (1 - y) (2 x - 1)$ $=>2(1+y)x(2x-1)+(1-y)(2x-1)$

$\Rightarrow (2 x - 1) (2 (1 + y) x + (1 - y)) = 0$ $=>(2x-1)(2(1+y)x+(1-y))=0$

Hence $x = \frac{1}{2}$ $x=1/2$

Checking

$4 (1 + y) x^{2} - 4 x y - (1 - y)$ $4(1+y)x^2-4xy-(1-y)$

$= 4 (1 + y) {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) y - (1 - y)$ $=4(1+y)(1/2)^2-4(1/2)y-(1-y)$

$= 1 + y - 2 y - 1 + y = 0$ $=1+y-2y-1+y=0$

The value of x such that 4(1+y)x2−4xy+1−y=04(1+y)x^2-4xy+1-y=0 is?