WAEC Past Questions and Answers - Page 1014

x\(^2\) + mx - n = 0

a = 1, b = m, c = -n

α + β = \(\frac{-b}{a}\) = \(\frac{-m}{1}\) = -m

αβ = \(\frac{c}{a}\) = \(\frac{-n}{1}\) = -n

the roots are = \(\frac{1}{α}\) and \(\frac{1}{β}\)

sum of the roots = \(\frac{1}{α}\) + \(\frac{1}{β}\)

\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{α+β}{αβ}\)

α + β = -m
αβ = -n

\(\frac{α+β}{αβ}\) = \(\frac{-m}{-n}\) → \(\frac{m}{n}\)

product of the roots = \(\frac{1}{α}\) * \(\frac{1}{β}\)

\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{1}{αβ}\) → \(\frac{1}{-n}\)

x\(^2\) - (sum of roots)x + (product of roots)
x\(^2\) - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx\(^2\) - mx - 1 = 0

F = m * a

d = t\(^2\) + 3t.

a = \(\frac{d^2d}{dt^2}\)

\(\frac{d[d]}{dt}\) = 2t + 3

\(\frac{d^2d}{dt^2}\) = 2m/s\(^2\)

a = 2m/s\(^2\)

F = m * a

F = 3 × 2 = 6N 

dy/dx = 2x - 6

y = ∫ 2x - 6

y = \(\frac{2x^2}{2} - 6 + c\)
y = x\(^2\) - 6x + c
passes through (1,2)
2 = 1\(^2\) - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x\(^2\) - 6x + c

y = x\(^2\) -  6x + 7

2x\(^2\) + 2y\(^2\) - x - 3y - 41

standard equation of circle
(x-a)\(^2\) + (x-b)\(^2\) = r\(^2\)
General form of equation of a circle.
x\(^2\) + y\(^2\) + 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x\(^2\) + 2y\(^2\) - x - 3y - 41

= x\(^2\) + y\(^2\) + 2gx + 2fy + c
2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0
divide through by 2

g = \(\frac{-1}{4}\) ; 2g = \(\frac{-1}{2}\)

f = \(\frac{-3}{4}\) ; 2f = \(\frac{-3}{2}\)

a = -g  → - \(\frac{-1}{4}\) ; = \(\frac{1}{4}\)

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))

No explanation has been provided for this answer.