Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 2x²− 7x + 3 = 0
Answers
SOLUTION :
Given : 2x² –7x + 3 = 0
On dividing the whole equation by 2,
(x² - 7x/2 + 3/2) = 0
Shift the constant term on RHS
x² - 7x/2 = - 3/2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of 7/2)² = (7/4)² both sides
x² - 7x/2 + (7/4)²= - 3/2 + (7/4)²
Write the LHS in the form of perfect square
(x - 7/4)² = - 3/2 + 49/16
[a² - 2ab + b² = (a - b)²]
(x - 7/4)² = (-3 × 8 + 49)/16
(x - 7/4)² = (-24 + 49)/16
(x - 7/4)² = 25/16
On taking square root on both sides
(x - 7/4) = √(25/16)
(x - 7/4) = ± 5/4
On shifting constant term (-7/4) to RHS
x = ± 5/4 + 7/4
x = 5/4 + 7/4
[Taking +ve sign]
x = (5 +7)/4
x = 12/4
x = 3
x = - 5/4 + 7/4
[Taking -ve sign]
x = (- 5 + 7)/4
x = 2/4
x = 1/2
Hence, the roots of the given equation are 3 & ½.
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Answer:
x
=
3
or
x
=
1
2
Explanation:
2
x
2
−
7
x
+
3
=
0
⇒
2
(
x
2
−
7
2
x
)
+
3
=
0
⇒
2
(
x
2
−
2
×
7
4
x
+
49
16
)
=
49
8
−
3
=
25
8
(
x
−
7
4
)
2
=
25
16
⇒
x
−
7
4
=
±
5
4
⇒
x
=
7
4
±
5
4
=
3
or
1
2