# How do you write a quadratic equation in standard form with solutions 2+3i, 2-3i?

Jun 21, 2016

${x}^{2} - 4 x + 13 = 0.$
Let the roots of the reqd. qudr. eqn. be $\alpha = 2 + 3 i , \beta = 2 - 3 i .$
Then, $\alpha + \beta = 4 ,$ &, $\alpha \cdot \beta = \left(2 + 3 i\right) \left(2 - 3 i\right) = 4 - 9 {i}^{2} = 4 + 9 = 13.$
Therefore, the reqd. qudr. eqn. is given by ${x}^{2} - \left(\alpha + \beta\right) x + \alpha \cdot \beta = 0 , .$ i.e., ${x}^{2} - 4 x + 13 = 0.$