# QUADRATIC WORD PROBLEMS - SOLVED EXAMPLES AND EXERCISE

if you have not already done so.

That knowledge is a prerequisite here.

### Example 1 :

Solve the Following Quadratic Word Problem.

The length and breadth of a rectangle differ by 17 m. The length and diagonal differ by 1 m. Find the length and breadth.

Solution to Example 1 :

Let l be the length of the rectangle. Then, by data it breadth = (l - 17).
Its diagonal = √{(l)2 + (l - 17)2}
By data, the length and diagonal differ by 1 m.
⇒ √{(l)2 + (l - 17)2} = l + 1Squaring both sides, we get
(l)2 + (l - 17)2 = (l + 1)2
l2 + l2 - 34l + 289 = l2 + 2l + 1
l2 - 36l + 288 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -36 and c = 288
We know by Quadratic Formula, l = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
l = {-b ± √(b2 - 4ac)}⁄2a
= [-(-36) ± √{ (-36)2 - 4(1)( 288)}]⁄2(1)= [36 ± √{ 36 x 36 - 4 x 8 x 36]⁄2 = [36 ± √{36(36 - 32)}]⁄2
= [36 ± √{36(4)}]⁄2 = [36 ± 6(2)}]⁄2 = 18 ± 6 = 18+6 or 18-6 = 24 or 12 Length can not be 12m,( because breadth = length - 17 = -5 is not possible.) ∴ l ≠ 12 ; ∴ l = 24.
Length = 24 m; Breadth = 24 - 17 = 7m. Ans.

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### Example 2 of Quadratic Word Problems

Solve the Following (Example 2 of Quadratic Word Problems ).

The length and breadth of a rectangle differ by 10 m. Its Area in square meters is 92 more than it perimeter in meters.Find the length and breadth.

Solution to Example 2 of Quadratic Word Problems :

Let l be the length of the rectangle. Then, by data, it breadth = (l - 10).
Its area = length x breadth = l(l - 10)
By data, Its Area in square meters is 92 more than it perimeter in meters.
l(l - 10) = 2 {l + (l - 10) } + 92 ⇒ l2 - 10l = 2(2l - 10) + 92 = 4l - 20 + 92 = 4l + 72
l2 - 14l - 72 = 0Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -14 and c = -72
We know by Quadratic Formula, l = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
l = {-b ± √(b2 - 4ac)}⁄2a
= [-(-14) ± √{(-14)2 - 4(1)(-72)}]⁄2(1)= [14 ± √{196 + 288}]⁄2 = [14 ± √{4(49 + 72)}]⁄2
= [14 ± √{4(121)}]⁄2 = [14 ± 2(11)}]⁄2 = 7 ± 11 = 7+11 or 7-11 = 18 or -4 Length can not be negative. ∴ l ≠ -4 ; ∴ l = 18.
Length = 18 m; Breadth = 18 - 10 = 8m. Ans.

### Example 3 of Quadratic Word Problems

Solve the Following (Example 3 of Quadratic Word Problems ).

The denominator of a fraction exceeds the numerator by 4and the fraction formed by squaring both numerator and denominatoris equal to 4⁄9. Find the fraction.

Solution to Example 3 of Example 2 of Quadratic Word Problems :

Let the denominator of the fraction be x. Then its numerator as per data is (x - 4)
Square of the fraction = {(x - 4) ⁄x}2 = 4⁄9 [By data ]
(x - 4)2x2 = 4⁄9
crossmultiplying, we get
9(x - 4)2 = 4x2 ⇒ 9(x2 - 8x + 16) = 4x2
⇒ 9x2 - 72x + 144 = 4x2 ⇒ 9x2 - 4x2 - 72x + 144 = 0
⇒ 5x2 - 72x + 144 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 5, b = -72 and c = 144
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-72) ± √{ (-72)2 - 4(5)( 144)}]⁄2(5)= [72 ± √{ 72 x 72 - 4(5)(2 x 72)}]⁄10 = [72 ± √{72 x 8(9 - 5)}]⁄10
= [72 ± √{8 x 9 x 8 x 4}]⁄10 = [72 ± 8 x 3 x 2)}]⁄10 = (72±48)⁄10 = (72+48)⁄10 or (72-48)⁄10 = 12 or 2.4
Taking the integral value, x = denominator = 12. Then Numerator = 12 - 4 = 8.
The required fraction = 8⁄12. Ans.

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### Example 4 of Quadratic Word Problems

Solve the Following (Example 4 of Quadratic Word Problems ).

A stream flows from A to B, a distance of 15 km. A man who can row in still water at 4 kmph can row up and down in 8 hours. Find the rate of the stream.

Solution to Example 4 of Quadratic Word Problems :

Let the rate of the stream be xkmph. xkmph;rowing speed of the man in the opposite direction of stream = rowing speed in still water - speed of the stream. = 4 kmph - xkmph;We know, Time = Distance⁄Speed
Time in hours for rowing up and down = (15 km)⁄( 4 + x)kmph + (15 km)⁄(4 - x)kmphBy data this is equal to 8 hours.
∴ 15⁄(4 + x) + 15⁄(4 - x) = 8
Multiplying both sides with (4 + x)(4 + x), we get
15(4 - x) + 15(4 + x) = 8(4 + x)(4 + x) ⇒ 15(4 - x + 4 + x) = 8(42 - x2)
⇒ 15(8) = 8(16 - x2) ⇒ 15 = 16 - x2
x2 = 16 - 15 = 1 ⇒ x = ± 1 = +1, -1
But x can not be negative. ∴ x = 1.
Thus, the rate of the stream = 1 kmph. Ans.

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### Exercise on Quadratic Word Problems

1. The product of two cosecutive integers is 812. Find them.
2. Thedifference of two numbers is 6 and their product is 391. Find them.
3. Twice the square of a number exceeds 4 times the number by 240. Find it.
For Answers See at the bottom of the Page.

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