# A rectangle has sides of (x + 3) \text{ and } (x - 4), what value of x gives an area of 78?

## Question:

A rectangle has sides of {eq}(x + 3) \text{ and } (x - 4), {/eq} what value of x gives an area of 78?

## Area of a Rectangle and Quadratic Equation:

(i) The area of a rectangle is {eq}\text{length} \times \text{width}. {/eq}

(ii) To solve a quadratic equation, we first solve the quadratic expression, set each factor equal to zero and solve the resultant equations.

The rectangle has sides of {eq}(x + 3) \text{ and } (x - 4) {/eq}.

So its area is given by:

\begin{align} \text{Area}&=(x+3) (x-4) \\ &= x^2-4x+3x-12 \\ &= x^2-x-12 \end{align}

But the area is given by 78.

So the above expression is equal to 78.

$$x^2-x-12=78 \\ \text{Subtracting 78 from both sides}, \\ x^2-x-90=0 \\$$

Here {eq}ac = 1 \times -90=-90 {/eq} and {eq}b=-1 {/eq}.

We have to find two numbers whose product is -90 and whose sum is -1.

Two such numbers are -10 and +9.

So the middle term {eq}-x {/eq} will be replaced by {eq}-10x+9x {/eq}.

Then the last equation becomes:

$$x^2-10x+9x-90=0 \\ (x^2-10x)+(9x-90)=0 \,\,\,\, \text{(Terms are grouped)} \\ x(x-10)+9(x-10)=0 \,\, \,\,\text{(Each group is factored)} \\ (x-10)(x+9)=0 \,\, \,\, \text{((x-10) is factored out)} \\ x-10=0; \,\, x+9=0 \\ x=10; \,\, x=-9$$

But {eq}x=-9 {/eq} would give one of the sides to be {eq}x-4=-9-4=-13 {/eq}, which is negative.

But the side of a rectangle cannot be negative.

Therefore, the answer is: {eq}\boxed{\mathbf{x=10}} {/eq}. 