how can there be roots to this equation (2 Viewers)

anonymoushehe · 2025-07-07T17:54:17+1000

delta is negative, so there are no real roots?

cheesynooby · 2025-07-07T18:12:38+1000

anonymoushehe said:
View attachment 48217
delta is negative, so there are no real roots?

there are no real roots but there are still... roots (complex roots, which is x2)
question should still be possible with only x1 content tho

Hehehe22 · 2025-07-07T18:41:07+1000

Well apparently you can still solve it using the angle formula for tan(alpha + beta), subbing in the sums and products of roots

liamkk112 · 2025-07-08T12:54:53+1000

anonymoushehe said:
View attachment 48217
delta is negative, so there are no real roots?

$\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$
in particular, you can factorise the equation like
$2(x-\tan(a))(x-\tan(b)) = 0 = 2x^2 -2x(\tan(a)+\tan(b))+2\tan(a)\tan(b)$
equating coefficients then gives you that
$\tan(a)+\tan(b) = \sqrt{3}/2, \tan(a)\tan(b) = 1/2$
(that's just sum/product of roots)
thus
$\tan(a+b) = \sqrt{3} \implies a+b = \arctan(\sqrt{3}) = \pi/3$

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how can there be roots to this equation (2 Viewers)

anonymoushehe

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cheesynooby

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Hehehe22

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liamkk112

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