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Solve for x:

x^4 + 10 x + 25 = 0

 

Subtract 10 x - 10 x^2 from both sides:

x^4 + 10 x^2 + 25 = 10 x^2 - 10 x

 

x^4 + 10 x^2 + 25 = (x^2 + 5)^2:

(x^2 + 5)^2 = 10 x^2 - 10 x

 

Add 2 (x^2 + 5) λ + λ^2 to both sides:

(x^2 + 5)^2 + 2 λ (x^2 + 5) + λ^2 = -10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2

 

(x^2 + 5)^2 + 2 λ (x^2 + 5) + λ^2 = (x^2 + 5 + λ)^2:

(x^2 + 5 + λ)^2 = -10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2

 

-10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2 = (2 λ + 10) x^2 - 10 x + 10 λ + λ^2:

(x^2 + 5 + λ)^2 = x^2 (2 λ + 10) - 10 x + 10 λ + λ^2

 

Complete the square on the right hand side:

(x^2 + 5 + λ)^2 = (x sqrt(2 λ + 10) - 5/sqrt(2 λ + 10))^2 + (4 (2 λ + 10) (λ^2 + 10 λ) - 100)/(4 (2 λ + 10))

 

Solve using the quadratic formula:

x = 1/2 (sqrt(2) sqrt(λ + 5) + sqrt(2) sqrt(-(25 + 10 λ + λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) or x = 1/2 (sqrt(2) sqrt(λ + 5) - sqrt(2) sqrt(-(25 + 10 λ + λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) or x = 1/2 (sqrt(2) sqrt((-25 - 10 λ - λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5)) - sqrt(2) sqrt(λ + 5)) or x = 1/2 (-sqrt(2) sqrt(λ + 5) - sqrt(2) sqrt((-25 - 10 λ - λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) where λ = -5 + (5 2^(2/3))/(3/5 (i sqrt(1119) + 9))^(1/3) + (5/6)^(2/3) (i sqrt(1119) + 9)^(1/3)

 

Substitute λ = -5 + (5 2^(2/3))/(3/5 (i sqrt(1119) + 9))^(1/3) + (5/6)^(2/3) (i sqrt(1119) + 9)^(1/3) and approximate:

x = -1.61762 - 1.035 i or x = -1.61762 + 1.035 i or x = 1.61762 - 2.04014 i or x = 1.61762 + 2.04014 i

Jan 26, 2018
 #1
avatar+33603 
+4
Jan 26, 2018
 #1
avatar+26364 
+1

Graph the image of the figure after a dilation with a scale factor of 2 centered at (−2, −2) .

 

 

\(\begin{array}{lrcll} \text{Formula dilation:} & \boxed{\vec{A}' = (\vec{A}-\vec{C_{Center}})\cdot \lambda +\vec{C_{Center}}} \\ & \vec{A} \text{ before dilation } & \text{$\lambda$ scale factor } = 2 \\ & \vec{A}' \text{ after dilation } & \text{$\vec{C_{Center}}$ center at $\binom{-2}{-2} $ } \\ \end{array} \)

 

\(\begin{array}{lrcll} &\vec{A}' &=& (\vec{A}-\vec{C_{Center}})\cdot \lambda +\vec{C_{Center}}\\ & &=& \vec{A}\cdot \lambda +\vec{C_{Center}}\cdot (1-\lambda ) \\ & &=& \lambda \cdot \vec{A} +(1-\lambda )\cdot \vec{C_{Center}} \\ \text{Formula dilation:} & \boxed{\vec{A}' =\lambda \cdot \vec{A} +(1-\lambda )\cdot \vec{C_{Center}}} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& (1-2)\binom{-2}{-2} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& (-1)\binom{-2}{-2} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& \binom{2}{2} \\ \end{array}\)

 

\(\begin{array}{lrcll} \boxed{\vec{A}'=\lambda \cdot \vec{A} +\binom{2}{2}}& \vec{A} = \dbinom{-4}{-2}\\ \boxed{\vec{B}'=\lambda \cdot \vec{B} +\binom{2}{2}}& \vec{B} = \dbinom{-5}{2}\\ \boxed{\vec{C}'=\lambda \cdot \vec{C} +\binom{2}{2}}& \vec{C} = \dbinom{-1}{3}\\ \boxed{\vec{D}'=\lambda \cdot \vec{D} +\binom{2}{2}}& \vec{D}= \dbinom{1}{1}\\ \end{array}\)

 

\(\begin{array}{lrcll} \boxed{\vec{A}'=2 \cdot \binom{-4}{-2} +\binom{2}{2}}& \vec{A}' = \dbinom{-6}{-2}\\ \boxed{\vec{B}'=2 \cdot \binom{-5}{2} +\binom{2}{2}}& \vec{B}' = \dbinom{-8}{6}\\ \boxed{\vec{C}'=2 \cdot \binom{-1}{3} +\binom{2}{2}}& \vec{C}' = \dbinom{0}{8}\\ \boxed{\vec{D}'=2 \cdot \binom{1}{1} +\binom{2}{2}}& \vec{D}' = \dbinom{4}{4}\\ \end{array} \)

 

Point (-4, -2 ) goes to (-6,-2)

Point (-5, 2) goes to (-8. 6)

Point ( -1, 3) goes to (0, 8)

Point ( 1, 1 ) goes to (4, 4 )

 

 

laugh

Jan 26, 2018

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