BXEECISE 21



Find tke real and complex roots of the following equations :



1. a;* - 8 0)2 + 8 a; + 15 = 0.
2. a;*-7a!2 + 12a; + 18 = 0.
3. a;*-4ar'-12a;-9 = 0.
4. a;* — 5ar — 10a; — 6 = 0.
5. a;*-5a;^-4a;-|-12 = 0.
6. a;* + 3 x^ + 6 a; - 10 =0.
7. «*-lla;2-14a; + 24=0.
8. a;^ - 10 a!^ + 20 a; - 16 = 0.
9. a;*-5a:'-l-10a;-6 = 0.
10. a;*-8a;2-|- 8 a; + 15 = 0.
75.* Solution of cubic equations by means of the standard
parabola.
To solve the cubic

a;3+6x2 + cx-l-(? = (1)

by means of a parabola, multiply by {x— 6), i.e. introduce the new root 5.
a;« + (c - J2)a;2 + ((j _ hc)x -bd = 0.
66



GBAPBIC ALGEBRA



Hence, applying § 70, we have







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A






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3'



6c — <?



. &2 _ c + 1



(2)
(3)



The formula for the radius is not neces-
sary, since the circumference must pass
through the point of the parabola whose
abscissa is 6, i.e. (6, 6^).
Thus, to solve

a;3 + 3 a;2 _ 6 a; - 8 = 0, (4)

either multiply by a; — 3, obtaining k* — 15z2•
+ 10 a; + 24 = 0, or apply directly formulae
(2) and (3).
Hence ai^ = — 5,

yo = 8.
From (—5, 8) as a center construct a
circle passing through A, i.e. the point in the
parabola whose abscissa is 3.
