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3 years ago

(x²+y²+2xy+1)dy=(x+y)dx

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

7 0

$( {x}^{2} + {y}^{2} + 2xy + 1)dy = (x + y)dx \\ \frac{dy}{dx} = \frac{x + y}{ {x}^{2} + {y}^{2} + 2xy + 1 }$

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Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

<u>Answer:</u>

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

<u>Solution:</u>

Need to solve given equation using quadratic formula.

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General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

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4 years ago

(x²+y²+2xy+1)dy=(x+y)dx​

(x²+y²+2xy+1)dy=(x+y)dx