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Find a quadratic equation in standard form with real coefficients that has the given solution . a.1-2i b.-2-i
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Start from the polynomial corresponding to the equation in factored form, that is let's say we know one solution is \(-3\) and the other is \(\sqrt{2}\), then we know \[ (x-(-3))(x-\sqrt{2})=0 \] Because if we put in the solutions we will get 0 in both cases. Now, if we expand the left hand side we get a quadratic equation with these solutions!
So in my example the quadratic formula with solutions \(-3\) and \(\sqrt{2}\) is \[ x^2+x(3-\sqrt{2})-3\sqrt{2}=0 \]
So how do I start with my problems?
Write the solutions as factors. A hint is that the first factor is \((x-(1-2i))\)
Okay now I have x,-1+2i
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