Question on roots of a quadratic equation (1 Viewer)

gr_111 · Jun 27, 2012

Need help with this question, Q8a from 2010 CSSA Trial Paper:

The quadratic equation $x^2 - 4Ax +6B = 0$ has two different real roots $\alpha$ and $\beta$ .

i) Show that $2A^2 > 3B$ .

ii) Find the range of values of $B$ if the sum of the roots is equal to the product of the roots.

Please show working, latex if possible. Thanks in advance.

asianese · Jun 27, 2012

For real roots, $\Delta>0$ so $(-4A)^2-4(1)(6B)>0\\\\ 16A^2>24B \\\\ 2A^2>3B$

If $\alpha+\beta = \alpha\beta$ then $-b/a = c/a \\\\ 4A=6B\\\\2A=3B$ But $2A^2>3B$ and $A=\frac{3B}{2}$ subbing into inequality, $2(\frac{3B}{2})^2>3B\\\\ \frac{2(9B^2)}{4}>3B\\\\ \frac{9B^2}{2}>3B$ QUADRATIC INEQUALITY!!

$9B^2-6B>0\\ 3B^2-2B>0 \\ B(3B-2)>0 \\ B<0, B>\frac{2}{3}$

gr_111 · Jun 27, 2012

Thanks, rep 4 u!

asianese · Jun 27, 2012

Question on roots of a quadratic equation (1 Viewer)

gr_111

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asianese

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gr_111

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asianese

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