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sqrt2x^2+sqrt3x+1=0 use the discriminant to determin how many real roots each equation
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that is not even right!
\[b^2-4ac=\sqrt{3}^2-4\times \sqrt{2}\times 1=3-4\sqrt{2}\]
\[b^2-4ac=(\sqrt{3})^2-4(\sqrt{2})(1)=3-4\sqrt{2}\]
ok is this the answer. This is what I I wrote for the equation
I guess i am confused
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no real roots the discriminant is <0
you job is to see if this number is positive, negative or zero. it is negative (easy check) so not real solutions
>0=> two real roots =0=> one real root (multiplicity 2) <0=> no real roots
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