Lawn dimensions from a quadratic area plan
A landscape contractor in Udaipur must lay a rectangular lawn in the courtyard of a Praveshika school. The plan gives the area of the lawn as (6x² + 11x + 3) square metres, where x is a positive number fixed by the width of the surrounding verandah. The mason has already marked one side of the lawn on the ground as (2x + 3) metres. The headmaster wants the second side written on the plan, and also the actual dimensions, because the verandah width has finally been fixed at x = 4. A 50-metre roll of fencing wire lies in the store for fencing the lawn on all four sides, and the headmaster claims the roll is just enough. The contractor disagrees, saying the perimeter grows faster than the area when x increases. The drawing teacher adds a third question: at what value of x would the planned area become zero, and can such a lawn actually be built?
1. Find the second side of the lawn by factorising 6x² + 11x + 3, and verify the product.
Read the worked answer
Split the middle term so that the two parts multiply to 6 × 3 = 18 and add to 11: the numbers are 9 and 2.
6x² + 11x + 3 = 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)
Since one side is (2x + 3) m, the second side is (3x + 1) m.
Verification: (2x + 3)(3x + 1) = 6x² + 2x + 9x + 3 = 6x² + 11x + 3, which matches the given area.
Note that both factors give positive lengths only when x > 0, which agrees with the statement that x is positive.
2. Taking x = 4, find the actual dimensions and the perimeter, and decide whether the 50 m wire is enough. Then test the contractor's claim that the perimeter grows faster than the area between x = 4 and x = 5.
Read the worked answer
At x = 4: sides are 2(4) + 3 = 11 m and 3(4) + 1 = 13 m.
Area = 11 × 13 = 143 m² (check: 6(16) + 44 + 3 = 143).
Perimeter = 2(11 + 13) = 48 m.
48 m < 50 m, so the roll is enough, with 2 m to spare - the headmaster is right at x = 4.
At x = 5: sides are 13 m and 16 m, perimeter = 58 m, area = 208 m².
Change from x = 4 to x = 5: area increases by 208 − 143 = 65 m², perimeter by 58 − 48 = 10 m. The area grows faster, so the contractor's statement is the reverse of the truth.
A practical rider: the 2 m spare wire disappears once x reaches 5, because 58 m > 50 m, so the stock position is safe only for the fixed value x = 4.
3. At what value of x does the planned area become zero, and can such a lawn be built?
Read the worked answer
Set the factorised area equal to zero:
(2x + 3)(3x + 1) = 0 gives x = −3/2 or x = −1/3.
Both roots are negative, but x represents a length-linked positive quantity (a verandah width), so neither value is admissible in this situation.
Conclusion: within the valid domain x > 0 the area 6x² + 11x + 3 is always positive - each factor is positive, so no permissible x makes the lawn vanish. The two negative roots are algebraic solutions of the equation only, and must be rejected on physical grounds. This is exactly the kind of check the board expects: state the roots, then justify rejection using the context.