Question

Solve the equation \[{{x}^{2}}+2x-63=0\] to find the perimeter of the rectangle taking the roots of the equation as dimensions of the rectangle

Answer 1

Hint: First of all factorize the given equation and find its roots or zeros. Then, use the positive part of its roots as length and breadth of the rectangle and use the formula 2 (l + b) to find the perimeter of the rectangle.

Complete step-by-step answer:
In this question, we are given a quadratic equation \[{{x}^{2}}+2x-63=0\] and we have to find the perimeter of the rectangle taking the roots of this equation as dimensions of the rectangle. Let us consider the quadratic equation given in the question.
\[{{x}^{2}}+2x-63=0\]
We can also write the above equation as
\[{{x}^{2}}+9x-7x-63=0\]
By grouping the terms of the above equation, we get,
\[\left( {{x}^{2}}+9x \right)-\left( 7x+63 \right)=0\]
By taking out the common terms from each group, we get,
\[x\left( x+9 \right)-7\left( x+9 \right)=0\]
By taking out (x + 9) common from the above equation, we get,
\[\left( x+9 \right)\left( x-7 \right)=0\]
So, from this we get, x = – 9 or x = 7.
Now, we know that the length and breadth of a rectangle cannot be negative. So, we take the length as 9 units and breadth as 7 units.

We know that the perimeter is the sum of all the sides of the polygon. So, we get the perimeter of the rectangle = 2(l + b). By substituting the value of l = 9 units and b = 7 units, we get,
Perimeter of the rectangle = 2 (9 + 7) = 2 (16) = 32 units.
So, we get the perimeter of the rectangle as 32 units.

Note: In this question, students can also find the roots of the equation by using the quadratic formula
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] and get the values of a, b, and c by comparing the given equation by standard quadratic equation which is \[a{{x}^{2}}+bx+c=0\]. Also, students must remember the formula for the perimeter of general polygons like square, rectangle, etc.