2011 - WAEC Mathematics Past Questions and Answers - page 3

x² - 3x = 7

x² - 3x - 7 = 0

What can you add to both sides of the equation to give the same value of y = x² - 3x - 3

The number is 4

x² - 3x - 7 + 4 = 4

x² - 3x - 3 = 4

but y = x² - 3x - 3

y = 4; So are y = 4 draw a line parallel to x axis, to cut or intersect the graph. At these points look down to see the corresponding values on x axis

This give -1.55 and 4.55

\(\frac{m}{n} + \frac{(m - 1)}{5n} - \frac{(m - 2)}{10n}\); \(\frac{10m + 2(m - 1) - 1(m - 2)}{10m}\)

= \(\frac{10m + 2m - 2 - m + 2}{10n}\)

= \(\frac{10m + 2m - m - 2 + 2}{10n}\)

= \(\frac{11m}{10n}\)

\(\sqrt{2} + \sqrt{32} - 3\sqrt{18} = x\sqrt{8}\)

= \(\sqrt{36 \times 2} + \sqrt{16 \times 2} - 3\sqrt{2 \times 9}\)

= x\(\sqrt{2 \times 4}\)

= 6\(\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = 2 \times \sqrt{2}\)

\(\sqrt{2} (6 + 4 - 9) = 2x\sqrt{2}\)

\(\sqrt{2} = 2x \sqrt{2}\) divide both sides by \(\sqrt{2}\)

\(\frac{\sqrt{2}}{\sqrt{2}} = \frac{2 \times \sqrt{2}}{\sqrt{2}}\)

1 = 2x

2x = 1

x = \(\frac{1}{2}\)

G \(\alpha\) H²

G = KH²

4 = K(3)²

4 = 9k; K = \(\frac{4}{9}\)

100 = \(\frac{4}{9}H^2\)

4H² = 900

H² = \(\frac{900}{4}\)

H² = 225

H = \(\sqrt{225}\)

H = 15

n(P \(\cup\) Q) = m(P \(\cap\) C)

38 = 19 = n(C) - 7

n(C) = 38 - 12

= 26

What must be added to 2x - 3y = Difference between x - 2y and 2x - 3y

x - 2y - 2y - (2x - 3y); x - 2y - 2x + 3y

= x - 2x + 3y - 2y

= y - x

1\(\frac{3}{4} - (2 \frac{1}{3} + 4)\) = \(\frac{7}{4} - (\frac{7}{3} + \frac{4}{1})\); \(\frac{7 + 12}{3}\)

\(\frac{7}{4} - \frac{19}{3} = \frac{21 + 76}{12}\)


= \(\frac{-55}{12} = -4 \frac{7}{12}\)

x² + 7x + 10 = 0

x² + 5x + 2x + 10 = 0

x(x + 5) + 2(x + 5) = 0

(x + 2)(x + 5) = 0

x + 2 = 0 or x + 5 = 0

x = -2 or x = -5

the smaller value x = -5

A = L x B

= (18.6 x 12.5)m²

= 232.5m²

= 233m²