1. Solve:

a) $x² - 9 = 0$

Solution:
Given,
$x² - 9 = 0$
$or, x² = 9$
$or, x = \pm \sqrt{9}$
$\therefore x = \pm 3


c) $\frac{9x²}{5} = \frac{5}{4}$

Solution:
Given,
$\dfrac{9x²}{5} = \dfrac{5}{4}$
$or, 4 × 9x² = 5×5$
$or, 36x² = 25$
$or, x² = \dfrac{25}{36}$
$or, x = \pm \sqrt{\frac{25}{36}}$
$\therefore x = \pm \dfrac{5}{6}$


e) $x² - 5x +6 = 0$

Solution:
Given,
$x² - 5x +6 = 0$
$or, x² - (2+3)x +6 = 0$
$or, x² -2x -3x +6 = 0$
$or, x(x -2) -3(x -2) = 0$
$or, (x -3)(x -2) = 0$
Either,
$x -3 = 0$
$\therefore x = 3$
Or,
$x -2 = 0$
$\therefore x = 2$
Hence, x = 2 or 3.


g) $x² - 2x -8 = 0$

Solution:
Given,
$x² -2x -8 = 0$
$or, x² - (4-2)x -8 = 0$
$or, x² -4x +2x -8 = 0$
$or, x(x -4) +2(x -4) = 0$
$or, (x +2)(x -4) = 0$
Either,
$x +2 = 0$
$\therefore x = -2$
Or,
$x -4 = 0$
$\therefore x = 4$
Hence, x = -2 or 4.


i) $5x² +8x -21 = 0$

Solution:
Given,
$5x² + 8x -21 = 0$
$or, 5x² + (15-7)x -21 = 0$
$or, 5x² + 15x -7x -21 = 0$
$or, 5x(x +3) - 7(x +3) = 0$
$or, (5x -7)(x +3) = 0$
Either,
$5x -7 = 0$
$or, 5x = 7$
$\therefore x = \frac{7}{5}$
Or,
$x +3 = 0$
$\therefore x = -3$
Hence, x = -3 or $\frac{7}{5}$.

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