Today our topic of discussion is Usage of quadratic equations.
Usage of quadratic equations
Usage of quadratic equations
Many problems of our daily life can be solved easily by forming linear and quadratic equations. Here, the formation of quadratic equations from the given conditions based on real life problems and techniques for solving them are discussed.
Example 12. Denominator of a proper fraction is 4 more than the numerator. If the fraction is squared, its denominator will be 40 more than the numerator. Find the fraction.
Solution: Let the numerator of the fraction bear and denominator x + 4 Hence the fraction is x/(x + 4)
Square of the fraction = (x/(x + 4))² = (x²)/((x + 4)²) = (x²)/(x² + 8x + 16) Here, numerator and denominator = x² + 8x + 16
by the question, x² + 8x + 16 = x² + 40
or, 8x + 16 = 40
or, 8x = 40 – 16
ot, 8x = 24 or, x = 3
x + 4 = 3 + 4 = 7 x/(x + 4) = 3/7
∴ the fraction is 3/7
Example 13. A rectangular garden with length 50 metre and breadth 40 metre has of equal width all around the inside of the garden. If the area of the garden except the path is 1200 square metre, how much is the path wide in metre?
Solution: Let the path be x metre wide
Without the path length of the garden is (50 – 2x) metre and breadth is (40 – 2x) metre
∴ Without the path area of the garden = (50 – 2x)(40 – 2x) square metre
By the question, (50 – 2x)(40 – 2x) = 1200
or, 2000 – 80x – 100x + 4x² = 1200
or, 4x² – 180x + 800 = 0
or, x² – 45x + 200 = 0[divi * dingby * 4]
ox, x² – 5x – 40x + 200 = 0
or, x(x – 5) – 40(x – 5) = 0 or, (x – 5)(x – 40) = 0
x – 5 = 0 or x – 40 = 0
if x – 5 = 0 5
if x – 40 = 0 40
But the breadth of the path will be less than 40 metre from the breadth of the garden.
therefore x ne40;
∴ x = 5
∴ the path is 5 metres wide
Example 14. Shahik bought some pens for Tk. 240. If he would get one more pen in that money, average cost of each pen would be less by Tk. 1. How many pens did he buy?
Solution: Let, shahik bought r pens in total for Tk. 240. Then each pen costs Tk . 240 x .
If he would get (x + 1) pens by Tk. 240, then the cost of each pen would be Tk. 240/(x + 1)
By the question, 240/(x + 1) = 240/x – 1
240/(x + 1) = (240 – x)/x or, 240x = (x + 1)(240 – x) by cross multiplication
or, x ² + x – 240 = 0 [by transposition]
or, 240x = 240x + 240 – x² – x
or, x² + 16x – 15x – 240 = 0 or, x(x + 16) – 15(x + 16) = 0
or, (x + 16)(x – 15) = 0
x + 16 = 0 or x – 15 = 0
If x + 16 = 0, x = – 16
If x – 15 = 0 15
But the number of pen z cannot be negative.
x ≠- 16 , x= 15
∴ Shahik bought 15 pens.
Example 15. In an examination of class IX of a school, total marks of r students obtained in mathematics is 1950. If at the same examination, marks of a new student in mathematics is 34 and it is added to the former total marks, the average of the marks become less by 1.
1) Write down the average of the obtained marks of all students including the new student and separately a students in terms of r.
2) By forming equation from the given conditions, show that, x²+ 35x – 1950 = 0
3) By finding the value of r, find the average of the marks in the two cases
Solution:
1) Average of the marks obtained by z students = 1950/x Average of the marks obtained by (x + 1) students including the new student = 195 overline 0 +34 x + 1 = 1984/(x + 1)
2) By the question. 1950/x = 1984/(x + 1) + 1
or, 1950/x – 1984/(x + 1) = 1 by transposition
or, (1950x + 1950 – 1984x)/(x(x + 1)) = 1
or, x² + x = 1950x – 1984x + 1950
or, x²+ x = 1950 – 34x
x² + 35x – 1950 = 0 [showed]
3) x² + 35x – 1950 = 0
0t – x ² + 65x – 30x – 1950 = 0
or, x(x + 65) – 30(x + 65) = 0 [ by cross multiplication]
or, (x + 65)(x – 30) = 0
x + 65 = 0 or x – 30 = 0
If 650, -65
Again, if x – 30 = 0 x = 30
Since the number of students, ie. x cannot be negative,
Hence, x ≠ -65
in the first case, average= 1950/31=65
and in the second case,average =1984/31=64
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