Notes: For what range of k values will the linear-quadratic system below have 2 solutions? y=2x^2+4+5 y=kx-3

Saturday, 21 June 2014

For what range of k values will the linear-quadratic system below have 2 solutions? y=2x^2+4+5 y=kx-3

So you have
$\displaystyle{2}{{x}}^{{2}}+{4}+{5}={k}{x}-{3}$

$\displaystyle{2}{{x}}^{{2}}-{k}{x}+{12}={0}$ this equation will have two solutions if and
only if its discriminant $\displaystyle{D}={{b}}^{{2}}-{4}{a}{c}\ne{q}{0}$ where $\displaystyle{a},{b},{c}$ are coeffitients of quadratic
equation.

$\displaystyle{{k}}^{{2}}-{4}\cdot{2}\cdot{12}\ne{q}{0}\Rightarrow{k}\ne{q}\pm\sqrt{{{96}}}$

So for $\displaystyle{k}=\pm\sqrt{{{96}}}$ you would have 1 solution and for all other values of $\displaystyle{k}$ you
would have 2 solutions which would be complex if discriminant is negative and real if
discriminant is positive.

Notes

Saturday, 21 June 2014

For what range of k values will the linear-quadratic system below have 2 solutions? y=2x^2+4+5 y=kx-3

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