+0

help

0
108
1

Find the product of all real numbers n which satisfy |n^2 - 9n + 20| = |16 - n^2|.

Nov 18, 2019

#1
+24407
+3

Find the product of all real numbers $$n$$ which satisfy $$|n^2 - 9n + 20| = |16 - n^2|$$.

$$\begin{array}{|lrcll|} \hline & \mathbf{|n^2 - 9n + 20|} &=& \mathbf{|16 - n^2|} \quad | \quad \text{square both sides} \\\\ & (n^2 - 9n + 20)^2 &=& (16 - n^2)^2 \quad | \quad n^2 - 9n + 20 =(n-5)(n-4) \\ & (n-5)^2(n-4)^2 &=& (16 - n^2)^2 \quad | \quad (16 - n^2) =(4-n)(4+n) \\ & (n-5)^2(n-4)^2 &=& (4-n)^2(4+n)^2 \\ & (n-5)^2(n-4)^2 - (4-n)^2(4+n)^2 &=& 0 \\ & \mathbf{(n-4)^2\Big( (n-5)^2 - (4+n)^2 \Big)} &=& \mathbf{0} \\ \hline 1. & \mathbf{(n-4)^2} &=& \mathbf{0} \\ & n-4 &=& 0 \\ & \mathbf{n_1} &=& \mathbf{4} \\ \hline 2. & \mathbf{(n-5)^2 - (4+n)^2} &=& \mathbf{0} \\ & (n-5)^2 &=& (4+n)^2 \\ & n-5 &=& \pm \sqrt{(4+n)^2 } \\ & \mathbf{n-5} &=& \mathbf{\pm (4+n)} \\\\ 3. & n-5 &=& 4+n \\ & -5 &\neq& 4 \quad | \quad \text{no solution} \\\\ 4. & n-5 &=& -(4+n) \\ & n-5 &=& -4 - n \\ & 2n &=& 1 \\ & \mathbf{n_2} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$$

The product of all real numbers $$n$$ is $$n_1\times n_2 = 4\times \dfrac{1}{2} = \mathbf{2}$$

Nov 18, 2019

#1
+24407
+3

Find the product of all real numbers $$n$$ which satisfy $$|n^2 - 9n + 20| = |16 - n^2|$$.

$$\begin{array}{|lrcll|} \hline & \mathbf{|n^2 - 9n + 20|} &=& \mathbf{|16 - n^2|} \quad | \quad \text{square both sides} \\\\ & (n^2 - 9n + 20)^2 &=& (16 - n^2)^2 \quad | \quad n^2 - 9n + 20 =(n-5)(n-4) \\ & (n-5)^2(n-4)^2 &=& (16 - n^2)^2 \quad | \quad (16 - n^2) =(4-n)(4+n) \\ & (n-5)^2(n-4)^2 &=& (4-n)^2(4+n)^2 \\ & (n-5)^2(n-4)^2 - (4-n)^2(4+n)^2 &=& 0 \\ & \mathbf{(n-4)^2\Big( (n-5)^2 - (4+n)^2 \Big)} &=& \mathbf{0} \\ \hline 1. & \mathbf{(n-4)^2} &=& \mathbf{0} \\ & n-4 &=& 0 \\ & \mathbf{n_1} &=& \mathbf{4} \\ \hline 2. & \mathbf{(n-5)^2 - (4+n)^2} &=& \mathbf{0} \\ & (n-5)^2 &=& (4+n)^2 \\ & n-5 &=& \pm \sqrt{(4+n)^2 } \\ & \mathbf{n-5} &=& \mathbf{\pm (4+n)} \\\\ 3. & n-5 &=& 4+n \\ & -5 &\neq& 4 \quad | \quad \text{no solution} \\\\ 4. & n-5 &=& -(4+n) \\ & n-5 &=& -4 - n \\ & 2n &=& 1 \\ & \mathbf{n_2} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$$

The product of all real numbers $$n$$ is $$n_1\times n_2 = 4\times \dfrac{1}{2} = \mathbf{2}$$

heureka Nov 18, 2019