Solve for y x=y^2+4y

1 answer:

4 0

$x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\$

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$x=y^2+4y\\ y^2+4y-x=0\\
y^2+4y+4-4-x=0\\
(y+2)^2=x+4\\
y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\$

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The right one is: 2x^3 + 4x^2 - 35x + 15.

So, the division is [2x^3 + 4x^2 - 35x + 15] / (x - 3)

The synthetic division uses the coeffcients and obviate the letters, but you have to be sure to respect the place of the coefficient.

So, in this case it is:

3 | 2      4    -35    15
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So, the quotient is 2x^2 + 10x - 5, and the remainder is 0.

I like to show it in this other way:

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Of course they are the same coefficients and the answer continue being quotien 2x^2 + 10x - 5, remainder 0.

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