Mathematics Questions and Answers

4x² + 5x + Q

To make a complete square, the coefficient of x² must be 1

= x² + \(\frac{5x}{4}\) + \(\frac{Q}{4}\)

Then (half the coefficient of x²) should be added

i.e. x² + \(\frac{5x}{4}\) + \(\frac{25}{64}\)

∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\)

Q = \(\frac{4 \times 25}{64}\)

= \(\frac{25}{16}\)

\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)

Let \(x^{-1}\) = a and \(y^{-1}\)= b

2a - 3b = 4 .......(i)

4a + b = 1 .........(ii)

(i) x 3 = 12a + 3b = 3........(iii)

2a - 3b = 4 ...........(i)

(i) + (iii) = 14a = 7

∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\)

From (i), 3b = 2a - 4

3b = 1 - 4

3b = -3

∴ b = -1

From substituting, \(2^{-1} = x^{-1}\)

∴ x = 2

\(y^{-1} = -1, y = -1\)

\(\frac{1}{2}\)(3y - 4x)² = (8x² + kxy + Ly²)

\(\frac{1}{2}\)(9y² - 24xy + 16x²) = 8x² + kxy + Ly²

\(\frac{9}{2}\)y² - 12xy) = kxy + Ly²

k = -12 ∴ L = \(\frac{9}{2}\)

4a² - 12ab - C² + 9b²

rearranges: (4a² - 12ab + 9b²) - c²

(2a - 3b)(2a - 3b) - c² = (2a - 3b)² - c²

= (2a - 3b + c)(2a - 3b - c)

\(y^3 -4xy + xy^3 - 4y \)

= \(y^3(1 + x) - 4y(1 + x)

\((y^3 - 4y)(1 + x) = (y^3(1 + x) - 4y(1 + x))\)

∴ = \(y(1 + x)(y + 2)(y - 2)\)