## Class 9 Mathematics Note

The quadratic equation in its standard form is ax2 + bx + c = 0

The value of x can be calculated as

x =$\frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}} {\rm{\: }}}}{{2{\rm{a}}}}$

Here, a, b, c are the real numbers such that a ≠ 0

Nature of roots

i. if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$> 0 ,it has two distinct and real roots

ii.if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$< 0 ,it has no real roots, roots are imaginary.

iii. if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$= 0 it has only one rootorrepeatedroots

Example: 1

X2 -16 = 0

X2 -16 = 0

X2 - 42 = 0

(x+4) (x- 4)= 0

x+4 = 0,i.e. x= -4

x- 4 = 0,i.e. x= 4

Therefore; x = ± 4

Example: 2

Solve 25x2-9 =0 by using formula.

25x2-9 =0

Or, 25x2 + O.x + (-9) = 0

Here, a=25

B=o

C=9

$\frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}$

$= \frac{{ - 0 \pm \sqrt {{{\rm{o}}^2} - 4{\rm{*}}25{\rm{*}}\left( { - 9} \right)} }}{{2{\rm{*}}25}}$

= ± $\frac{{30}}{{50}}$

=± $\frac{3}{5}$

Example: 3

The area of the rectangular room is 45sq. cm if the length had been 3m less and breadth 1m , more it would have been a square . Find the length and breadth of room.

Or, l*b = 45

So, l = $\frac{{45}}{{\rm{b}}}$….(i)

Or, l – 3 = b + 1

Or, l – b = 4….(ii)

Or, $\frac{{45}}{{\rm{b}}} - {\rm{b}} = 4$

Or, $\frac{{45 - {{\rm{b}}^2}}}{{\rm{b}}} = 4$

Or, 45 – b2 = 4b

Or, b2 + 4b – 45 = 0

Or, b2 + 9b – 5b – 45 = 0

Or, b(b – 5) + 9(b – 5) = 0

Or, (b – 5)(b + 9) = 0

So, b = 5, -9.

So, b = 5cm

Now,

Or, l – b = 4

Or, l – 5 = 4

So, l = 9cm

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