Mathematics Questions and Answers

y = 2x² + 9x - 35

2x² + 9x = 35

x² + \(\frac{9}{2}\) = \(\frac{35}{2}\)

x² + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\)

(x + \(\frac{9}{4}\))² = \(\frac{361}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{19}{4}\)

= 2.5 or -7

-7 < x < \(\frac{5}{2}\)

Let the present ages of father be x and son = y

five years ago, father = x - 5

son = y - 5

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -15 + 5 = -10 ......(i)

x + y = 110......(ii)

eqn(ii) - eqn(i)

4y = 120

y = 30

sub for y = 30 in eqn(i)

x -10 = 3(30)

x -10 = 90

x = 80

\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)

N : b Look for the antilog of .3482 which gives 222.9

x\(^2\)  + 4 = 0

x\(^2\)  = -4

You can't apply a square root to a negative number. So the answer is None of the above.

\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\)

Squaring both sides,

\(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\)

Equating the rational and irrational parts, we have

\(a + b = 170 ... (1)\)

\(2 \sqrt{ab} = 20 \sqrt{30}\)

\(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \)

\(ab = 3000 ... (2)\)

From (2), \(b = \frac{3000}{a}\)

\(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\)

\(a^{2} - 170a + 3000 = 0\)

\(a^{2} - 20a - 150a + 3000 = 0\)

\(a(a - 20) - 150(a - 20) = 0\)

\(\text{a = 20 or a = 150}\)

\(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\)

\(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\)

= \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)