Find the real solution of the equation:
1. x^2+2x-15 = 0
2. 2y^2-y-(1/2) = 0
**Note. (1/2) is in fraction. LOL. :)
3. x^2 - sqrt5 x +1 = 0
4. 3x^2+2x+2=0
5. 25x^2+70x+49=0
+130511 1. x^2+2x-15 = 0 factor this
(x + 5) (x - 3) = 0 and setting both factors to 0, we have
x = -5 or x = 3
2. 2y^2 - y - 1/2 = 0 use the quadratic formula for this one.....we have
$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{y}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.309\: \!016\: \!994\: \!374\: \!947\: \!4}}\\
{\mathtt{y}} = {\mathtt{0.809\: \!016\: \!994\: \!374\: \!947\: \!4}}\\
\end{array} \right\}$$
3. x^2 - sqrt5 x +1 = 0 this one is also a good candidate for the quad formula
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$
BTW....this is an interesting solution......the first answer is "phi" and the second is "Phi"
4. 3x^2+2x+2=0 this has no real solution because the discriminant is < 0.... (b^2 - 4ac) = (2^2 -4*2*3) = (4 -24)= -20
5. 25x^2+70x+49=0 this is a perfect square trinomial. so we have
(5x + 7) (5x + 7) = 0 and setting either factor to 0, we have that x = -7/5
+130511 1. x^2+2x-15 = 0 factor this
(x + 5) (x - 3) = 0 and setting both factors to 0, we have
x = -5 or x = 3
2. 2y^2 - y - 1/2 = 0 use the quadratic formula for this one.....we have
$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{y}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.309\: \!016\: \!994\: \!374\: \!947\: \!4}}\\
{\mathtt{y}} = {\mathtt{0.809\: \!016\: \!994\: \!374\: \!947\: \!4}}\\
\end{array} \right\}$$
3. x^2 - sqrt5 x +1 = 0 this one is also a good candidate for the quad formula
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$
BTW....this is an interesting solution......the first answer is "phi" and the second is "Phi"
4. 3x^2+2x+2=0 this has no real solution because the discriminant is < 0.... (b^2 - 4ac) = (2^2 -4*2*3) = (4 -24)= -20
5. 25x^2+70x+49=0 this is a perfect square trinomial. so we have
(5x + 7) (5x + 7) = 0 and setting either factor to 0, we have that x = -7/5
