2° Functions, Delta

ax^2 +bx +c has 2 solutions, x1 and x2  x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? YES

$\small{ \begin{array}{lrcll} (1) & x_1+x_2 &=& k & \qquad \rightarrow \qquad x_2 = k-x_1\\ (2) & \frac{x_1}{x_2} &=& k & \qquad \rightarrow \qquad x_1 = k \cdot x_2 \qquad \rightarrow \qquad x_1 = k \cdot (k-x_1)\\ \\ \hline \\ (2) & x_1 &=& k \cdot (k-x_1) \\ & x_1 &=& k^2 -k \cdot x_1 \\ & x_1 +k \cdot x_1 &=& k^2 \\ & x_1(1 +k ) &=& k^2 \\ & \mathbf{x_1} & \mathbf{=} & \mathbf{ \frac{k^2}{1 +k} }\\ \\ \hline \\ (1) &\frac{x_1}{x_2} &=& k \\ & x_2 &=& \frac{x_1}{k} \\ & x_2 &=& \frac{\frac{k^2}{1 +k}}{k} \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{ \frac{k}{1 +k} } \\ \\ \hline \end{array} }$

$\small{ \begin{array}{rcll} (x-x_1)(x-x_2) &=& 0 \\ (x - \frac{k^2}{ 1 + k } )(x - \frac{k}{ 1 + k } ) &=& 0 \\ x^2-x\cdot( \frac{k}{ 1+k } + \frac{k^2}{ 1+k } ) + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot[ \frac{k}{ 1+k }( 1+k ) ] + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot k + \frac{k^3}{ (1+k)^2 } &=& 0 \qquad | \qquad \cdot(1+k)^2 \\ \underbrace{(1+k)^2 }_{=a}\cdot x^2 \underbrace{- k \cdot (1+k)^2 }_{=b}\cdot x + \underbrace{k^3}_{=c} &=& 0 \\ \\ \hline \\ \end{array} }$

Example 1:

$\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 1 \\ a & = & (1+1)^2 = 4 \\ b & = & -1\cdot (1+1)^2 = -4\\ c & = & 1^3 = 1\\\\ 4x^2-4x+1 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {4 \pm \sqrt{(-4)^2-4\cdot 4\cdot 1} \over 2\cdot 4} \\ x_{1,2} &=& {4 \pm \sqrt{0} \over 8 } \\ x_{1,2} &=& \frac{4}{8} \\ x_{1,2} &=& \frac{1}{2} \\ x_1 &=& \frac{1}{2} \text{ or } x_1 &=& \frac{1}{2} \\\\ x_1+x_2 &=& \frac{1}{2}+\frac{1}{2} = 1 \\ \frac{x_1}{x_2} &=& \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \end{array}$

Example 2:

$\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 2 \\ a & = & (1+2)^2 = 9 \\ b & = & -2\cdot (1+2)^2 = -18\\ c & = & 2^3 = 8\\\\ 9x^2-18x+8 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {18 \pm \sqrt{(-18)^2-4\cdot 9\cdot 8} \over 2\cdot 9} \\ x_{1,2} &=& {18 \pm \sqrt{324-288} \over 18 } \\ x_{1,2} &=& {18 \pm \sqrt{36} \over 18 } \\ x_{1,2} &=& {18 \pm 6 \over 18 } \\ x_1 = \frac{18+6}{18} &\text{ or }& x_1 = \frac{18-6}{18} \\ x_1 = \frac{24}{18} &\text{ or }& x_1 = \frac{12}{18} \\\\ x_1 = \frac{4}{3} &\text{ or }& x_1 = \frac{2}{3} \\\\ x_1+x_2 &=& \frac{4}{3}+\frac{2}{3} = \frac{6}{3} = 2 \\ \frac{x_1}{x_2} &=& \frac{\frac{4}{3}}{\frac{2}{3}} = \frac{4}{2} = 2 \end{array}$

If $\displaystyle x_{1}+x_{2}=x_{1}/x_{2},$

then, solving for  $x_{1}$ ,

$\displaystyle x_{1}=\frac{x^{2}_{2}}{(1-x_{2})}$.

Now choose a value for x2, anything other than 1, and calculate the corresponding value of x1.

For example x2 = 2 gets you x1 = -4, and the quadratic

$\displaystyle (x-2)(x+4)=x^{2}+2x-8=0.$

There's an infinite number of possibilities.

ax^2 +bx +c has 2 solutions, x1 and x2  x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? YES

New edit, without mistake:

$\small{ \begin{array}{lrcll} (1) & x_1+x_2 &=& k & \qquad \rightarrow \qquad x_2 = k-x_1\\ (2) & \frac{x_1}{x_2} &=& k & \qquad \rightarrow \qquad x_1 = k \cdot x_2 \qquad \rightarrow \qquad x_1 = k \cdot (k-x_1)\\ \\ \hline \\ (2) & x_1 &=& k \cdot (k-x_1) \\ & x_1 &=& k^2 -k \cdot x_1 \\ & x_1 +k \cdot x_1 &=& k^2 \\ & x_1(1 +k ) &=& k^2 \\ & \mathbf{x_1} & \mathbf{=} & \mathbf{ \frac{k^2}{1 +k} }\\ \\ \hline \\ (1) &\frac{x_1}{x_2} &=& k \\ & x_2 &=& \frac{x_1}{k} \\ & x_2 &=& \frac{\frac{k^2}{1 +k}}{k} \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{ \frac{k}{1 +k} } \\ \\ \hline \end{array} }\\ \small{ \begin{array}{rcll} (x-x_1)(x-x_2) &=& 0 \\ (x - \frac{k^2}{ 1 + k } )(x - \frac{k}{ 1 + k } ) &=& 0 \\ x^2-x\cdot( \frac{k}{ 1+k } + \frac{k^2}{ 1+k } ) + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot[ \frac{k}{ 1+k }( 1+k ) ] + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot k + \frac{k^3}{ (1+k)^2 } &=& 0 \qquad | \qquad \cdot(1+k)^2 \\ \underbrace{(1+k)^2 }_{=a}\cdot x^2 \underbrace{- k \cdot (1+k)^2 }_{=b}\cdot x + \underbrace{k^3}_{=c} &=& 0 \\ \\ \hline \\ \end{array} }$

Example 1:

$\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 1 \\ a & = & (1+1)^2 = 4 \\ b & = & -1\cdot (1+1)^2 = -4\\ c & = & 1^3 = 1\\\\ 4x^2-4x+1 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {4 \pm \sqrt{(-4)^2-4\cdot 4\cdot 1} \over 2\cdot 4} \\ x_{1,2} &=& {4 \pm \sqrt{0} \over 8 } \\ x_{1,2} &=& \frac{4}{8} \\ x_{1,2} &=& \frac{1}{2} \\ x_1 = \frac{1}{2} &\text{ or }& x_2 = \frac{1}{2} \\\\ x_1+x_2 &=& \frac{1}{2}+\frac{1}{2} = 1 \\ \frac{x_1}{x_2} &=& \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \end{array}$

Example 2:

$\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 2 \\ a & = & (1+2)^2 = 9 \\ b & = & -2\cdot (1+2)^2 = -18\\ c & = & 2^3 = 8\\\\ 9x^2-18x+8 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {18 \pm \sqrt{(-18)^2-4\cdot 9\cdot 8} \over 2\cdot 9} \\ x_{1,2} &=& {18 \pm \sqrt{324-288} \over 18 } \\ x_{1,2} &=& {18 \pm \sqrt{36} \over 18 } \\ x_{1,2} &=& {18 \pm 6 \over 18 } \\ x_1 = \frac{18+6}{18} &\text{ or }& x_2 = \frac{18-6}{18} \\ x_1 = \frac{24}{18} &\text{ or }& x_2 = \frac{12}{18} \\\\ x_1 = \frac{4}{3} &\text{ or }& x_2 = \frac{2}{3} \\\\ x_1+x_2 &=& \frac{4}{3}+\frac{2}{3} = \frac{6}{3} = 2 \\ \frac{x_1}{x_2} &=& \frac{\frac{4}{3}}{\frac{2}{3}} = \frac{4}{2} = 2 \end{array}$