Mathematics Questions and Answers

x³ + px² + qx + 1 = (x - 1) Q(x) + R

x - 2 = 0, x = 2, R = 0,

4p + 2p = -9........(i)

x³ + px² + qx + 1 = (x - 1)Q(x) + R

-1 + p - q + 1 = 0

p - q = 0.......(ii)

Solve the equation simultaneously

p = \(\frac{-3}{2}\)

q = \(\frac{-3}{2}\)

p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\)

= \(\frac{-6}{2}\)

= -3

4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b)

8a3 + 27b3 = (2a)3 + (3b)3

= (2a + 3b)(4a - 6ab = 9a2)

(4a + 6b)2 = 2(2a + 3b)2

1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\)

sum of roots - \(1 + \sqrt{13} + 1 - \sqrt{13} = 2\)

Product of roots = (1 - \(\sqrt{13}\)) (1 + \(\sqrt{13}\)) = -12

x2 - (sum of roots) x + (product of roots) = 0

x2 - 2x - 12 = 0

f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4

= (2 + 3) (x - 3) + 4

= 5(x - 3) + 4

= 5x - 15 + 4

= 5x - 11

f(3) = 5 x 3 - 11

= 4

g(f(3)) = g(4)

= \(\sqrt{5 + 4}\)

= \(\sqrt{9}\)

= 3

g(4) = 3

f(g(4)) = f(3)

= 4

g[f(3)] and f[g(4)] = 3 and 4 respectively.

Tunde and Shola can do the work in 18 days.

Both will do the work in \(\frac{1}{18}\) days.

But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day.

Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day.

\(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\)

\(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\)

\(x^{2} + 15x = 36x + 270\)

\(x^{2} + 15x - 36x - 270 = 0\)

\(x^{2} - 21x - 270 = 0\)