Solve - Quadratic Equation | SciPiPupil

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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