Question

If a rectangle has an area of 6x² + 9x – 27, what are its dimensions?

Answer 1

Answer:

$0= {6x}^{2} + 9x - 27 \\ 0 = 3( {2x}^{2} + 3x - 9) \\ 0 = {2x}^{2} + 3x - 9 \\ 0 = {2x}^{2} + 6x - 3x - 9 \\ 0 = 2x(x + 3) - 3(x + 3) \\ 0 = (2x - 3)(x + 3) \\ x + 3 = 0 \: or \: 2x - 3 = 0 \\ x = - 3 \: or \: x = \frac{3}{2}$

Answer 2

Given:

• Area of the rectangle is 6x² + 9x – 27

To find:

• Dimensions of the rectangle ?

Solution:

• Let's consider ABCD is a rectangle.

Where,

• Area = 6x² + 9x – 27

• Let's consider length & breadth be l & b

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« Now, Finding the dimensions of the rectangle,

→ Area = l × b

In order to find the length and breadth of the rectangle respectively, we have to factorise the sum,

→ 6x² + 9x - 27 (18 × (-9) = (-27) × 6)

→ 6x² + 18x - 9x - 27

→ 6x(x + 3) - 9(x+3)

→ (6x - 9)(x+3)

∴ Hence, (6x - 9) , (x+3) are the dimensions of the rectangle.

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« Now, let's verify this,

L × B = 6x² + 9x - 27

→ (6x - 9) × (x + 3) = 6x² + 9x - 27

→ 6x² + 18x - 9x - 27 = 6x² + 9x - 27

→ 6x² + 9x - 27 = 6x² + 9x - 27

LHS = RHS

• Hence, Verified.