Question

$\text{The quadratic equation}{x}^2+7x+12=0\text{can be visualised to be a rectangle}\text{of width}(x+3)\text{and length}$ .

• A. ($x+3$)
• B. ($x+4$)
• C. ($x+1$)
• D. ($x+2$)

Answer 1

The correct option is B ($x+4$)

Let us visualise the quadratic equation $x^2+7x+12=0$ as a rectangle.

For this, the equation has to be expressed as the product of the width ($x+3$) and length of the rectangle.

Comparing the given equation to the standard form $a{x}^2+bx+c,$ where $a,b$ and $c$ are constants ($a\ne 0$) and factorising $a\times c$ such that the numbers add up to $b,$ we have
$b=7=4+3.$

$\therefore x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)$

$\therefore \text{Area of the rectangle}=(x+4)(x+3)=x^2+7x+12$

If $(x+3)$ is the width of the rectangle, then its length is $(x+4).$