Module-1 Algebra Quadratic Equations Worksheet With Answers Page 5
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14MODULE -
1
Quadratic Equations
Algebra
3
1
=
−
x =
x
Therefore,
and
are solutions of the given equation.
2
3
2
Example 6.5:
Solve x
+ 2x + 1 = 0
Notes
2
Solution:
We have
x
+ 2x + 1 = 0
2
or
(x + 1)
= 0
or
x + 1 = 0
which gives
x = – 1
Therefore, x = –1 is the only solution.
Note: In Examples 6.3 and 6.4, you saw that equations had two distinct solutions. However,
in Example 6.5, you got only one solution. We say that it has two solutions and these are
coincident.
CHECK YOUR PROGRESS 6.3
1. Solve the following equations using factor method.
2
(i) (2x + 3) (x + 2) = 0
(ii) x
+ 3x – 18 = 0
2
2
(iii) 3x
– 4x – 7 =0
(iv) x
– 5x – 6 = 0
(v) 25x
2
– 10x + 1 = 0
(vi) 4x
2
– 8x + 3 = 0
Quadratic Formula
Now you will learn to find a formula to find the solution of a quadratic equation. For this,
2
we will rewrite the general quadratic equation ax
+ bx + c = 0 by completing the square.
2
We have ax
+ bx + c = 0
2
Multiplying both sides by '4a' to make the coefficient of x
a perfect square, of an even
number, we get
2
2
4a
x
+ 4abx + 4ac = 0
2
2
2
2
or
(2ax)
+ 2(2ax)b + (b)
+ 4ac = b
[adding b
to both sides]
2
2
2
or
(2ax)
+ 2(2ax)b + (b)
= b
– 4ac
{
}
(
)
2
+
2
=
±
−
or
2
2ax
b
b
4ac
+
=
±
−
or
2
2ax
b
b
4ac
174
Mathematics Secondary Course
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