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Preguntas 11
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 #1
avatar+19496 
+2

Advanced Quadratic

Find the product of the \(y\)-coordinates of all the distinct solutions \((x,y)\) for the two equations
\(y=x^2-8\) and \(y^2=-5x+44\).

 

\(\begin{array}{|lrcll|} \hline (1) & y &=& x^2-8 \\\\ (2) & y^2 &=& -5x+44 \quad | \quad y^2 = (x^2-8)^2 \\ & (x^2-8)^2 &=& -5x+44 \\ & x^4-16x^2+84 &=& -5x+44 \\ & x^4-16x^2+5x+20 &=& 0 \\ & x^2(x^2-16)+5(x+4) &=& 0 \quad | \quad x^2-16 = (x-4)(x+4) \\ & x^2(x-4)(x+4)+5(x+4) &=& 0 \\ & (x+4)\left(x^2(x-4)+5\right) &=& 0 \\ \\ 1. & x_1+4 &=& 0 \\ & \mathbf{x_1} &\mathbf{=}& \mathbf{-4} \\ & y_1 &=& x_1^2-8 \\ & y_1 &=& (-4)^2-8 \\ & \mathbf{y_1} &\mathbf{=}& \mathbf{8} \\ & \text{solution $(-4,8)$ } \\\\ 2. & x^2(x-4)+5 &=& 0 \\ & x^3-4x^2 + 5 &=& 0 \\ \text{try } x=-1 & (-1)^3-4(-1)^2 + 5 &\overset{?}{=}& 0 \\ & -1-4 + 5 & = & 0\ \checkmark \\ & \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\ & y_2 &=& x_2^2-8 \\ & y_2 &=& (-1)^2-8 \\ & \mathbf{y_2} &\mathbf{=}& \mathbf{-7} \\ & \text{solution $(-1,-7)$ } \\\\ \hline \end{array}\)

 

Polynomial long division

\(\begin{array}{|rcll|} \hline x^2-5x+5 &=& 0 \\ x &=& \dfrac{5\pm\sqrt{25-4\cdot 5 }}{2}\\ x &=& \dfrac{5\pm\sqrt{5}}{2} \\\\ \mathbf{x_3} &\mathbf{=}& \mathbf{\dfrac{5+\sqrt{5}}{2}=3.618 } \\ y_3 &=& x_3^2-8 \\ y_3 &=& \left(\dfrac{5+\sqrt{5}}{2} \right)^2-8 \\ \mathbf{y_3} &\mathbf{=}& \mathbf{\dfrac{5\sqrt{5}-1}{2} = 5.09 } \\ \text{solution $(\dfrac{5+\sqrt{5}}{2},\dfrac{5\sqrt{5}-1}{2}) \\ =(3.618,5.09)$ } \\\\ \mathbf{x_4} &\mathbf{=}& \mathbf{\dfrac{5-\sqrt{5}}{2}=1.382 } \\ y_4 &=& x_4^2-8 \\ y_4 &=& \left(\dfrac{5-\sqrt{5}}{2} \right)^2-8 \\ \mathbf{y_4} &\mathbf{=}& \mathbf{-\left(\dfrac{5\sqrt{5}+1}{2}\right) = -6.09 } \\ \text{solution $(\dfrac{5-\sqrt{5}}{2},-\left(\dfrac{5\sqrt{5}+1}{2}\right)) \\ =(1.382,-6.09)$ } \\ \hline \end{array}\)

 

 

Find the product of the y-coordinates of all the distinct solutions (x,y):

\(\begin{array}{|rcll|} \hline && y_1y_2y_3y_4 \\ &=& 8\times(-7)\times \left(\dfrac{5\sqrt{5}-1}{2}\right) \times \left(-\left(\dfrac{5\sqrt{5}+1}{2}\right)\right) \\ &=& 8\times(-7)\times (-31) \\ &\mathbf{=}& \mathbf{1736} \\ \hline \end{array}\)

 

laugh

heureka 22-jun-2018
 #1
avatar+19496 
+3

10 sin^2(x) + 10 sin (x) cos (x) - cos^2 (x) = 2.

Solve for all values of  x between 0 and 360 degrees

 

\(\begin{array}{|rcll|} \hline 10 \sin^2(x) + 10 \sin (x) \cos (x) - \cos^2 (x) &=& 2 \\ && \small{\boxed{\sin(x) = \tan(x)\cos(x)}} \\ 10 [\tan(x)\cos(x)]^2 + 10 \tan(x)\cos(x)\cos (x) - \cos^2 (x) &=& 2 \\ 10 \tan^2(x)\cos^2(x) + 10 \tan(x)\cos^2(x) - \cos^2 (x) &=& 2 \\ \cos^2(x) \Big( 10 \tan^2(x) + 10 \tan(x) - 1 \Big) &=& 2 \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot \dfrac{1}{\cos^2(x) } \\ && \small{\boxed{ \dfrac{1}{\cos^2(x) } = 1+\tan^2(x)} } \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot( 1+\tan^2(x)) \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 +2\tan^2(x) \\ 8 \tan^2(x) + 10 \tan(x) - 3 &=& 0 \\ && \large{\boxed{\tan(x) = z}} \\ 8 z^2 + 10z - 3 &=& 0 \\\\ z&=& \frac{-10\pm \sqrt{100-4\cdot 8\cdot(-3)} } {2\cdot 8} \\ z&=& \frac{-10\pm \sqrt{196} } {16} \\ z&=& \frac{-10\pm 14 } {16} \\\\ z_1 &=& \dfrac{-10 + 14 } {16} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\dfrac{1} {4}} \\\\ z_2 &=& \dfrac{-10 - 14 } {16} \\ \mathbf{z_2} &\mathbf{=}& -\mathbf{\dfrac{3} {2}} \\ \hline \end{array}\)

 

solutions:

\(\begin{array}{|rcll|} \hline \tan(x) &=& z_1 \\ \tan(x) &=& \frac{1}{4} \\ \mathbf{x} &\mathbf{=}& \mathbf{\arctan(\frac{1}{4}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \tan(x) &=& z_2 \\ \tan(x) &=& -\frac{3}{2} \\ x &=& \arctan(-\frac{3}{2}) + n\cdot 180^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{-\arctan(\frac{3}{2}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array}\)

 

 x between 0 and \( \mathbf{360^{\circ}}\)

\(\begin{array}{|rcll|} \hline x_1 &=& \arctan(\frac{1}{4}) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{14.0362434679^{\circ}} \\\\ x_2 &=& -\arctan(\frac{3}{2}) + 180^{\circ} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{123.690067526^{\circ}} \\\\ x_3 &=& \arctan(\frac{1}{4})+ 180^{\circ} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{194.036243468^{\circ}} \\\\ x_4 &=& -\arctan(\frac{3}{2}) + 2\cdot 180^{\circ} \\ \mathbf{x_4} &\mathbf{=}& \mathbf{303.690067526^{\circ}} \\ \hline \end{array}\)

 

 

laugh

heureka 18-jun-2018