QUADRATIC EQUATIONS
(2x + 3)
+ 3x
– 90
Sol.
Let the number of articles produced in a day be x
A cottage industry produces a certain number of pottery articles in a day. It was
observed on a particular day that the cost of production of each article(in rupees)
was 3 more than twice the number of articles produced on that day. If the total
cost of production on that day was Rs.90, find the number of articles produced
and the cost of each article.
Total cost of
production
+ 3
=
Number
of articles
cost of each
article
×
90 =
x
90 =
2x2
=
0
∴
90
2x2
+
3x
‒
=
0
0
∴
12x
2x2
+
15x
‒
90
‒
=
∴
6
x
(2x + 15)
‒
(2x +15)
=
∴
(2x +15)
(x - 6)
0
=
∴
2x +15 = 0
or
x - 6
0
=
∴
2x +15 = 0
or
x - 6
0
=
∴
2x = -15
or
x = 6
No. of articles produced cannot be negative
∴
x
≠
-15
2
∴
x = 6
∴
2(6)
2x
+
3
=
3
+
=
12
3
+
=
15
x =
-15
2
or
x = 6
6 articles are produced each day and cost of production of each article is Rs.15.
Q.
2x
It is given that …
Calculation
180
90
45
2
2
3
15
3
5
5
1
Find two factors of 180 in such a way that by subtracting factors we get middle number.
‘-’ sign means subtracting
Let us do the prime factorization of 180
In a comparative statement whatever comes later is taken as ‘x’
What we have to find in this sum ?
∴
Since last sign is ‘-’ Give middle sign to the bigger factor & opposite sign to smaller factor.
15
12
180
+
–
15 - 12
=
3
90 × 2 = 180
+ 3x
0 =
2x2
EX 4.2 6