Question

if the sum of the square of the number and four times a number is 21 what is the number​

Answer 1

Given :-

Sum of the square of the number and four times a number = 21

To Find :-

• The number

Assumption :-

Let the number be x

Solution :-

According to question,

x² + 4x = 21

⟹ x² + 4x - 21 = 0

⟹ x² + (7 - 3x) - 21 = 0

⟹ x² + 7x - 3x - 21 = 0

⟹ x (x + 7) - 3 (x + 7)

⟹ (x - 3) (x + 7)

Thus,

x = 3 or -7

✰ Required Answer :-

The number = 3 or -7

Answer 2

Given: Sum of the square of the number and four times a number = 21

To Find: The number

Solution:

 Let the number be $x.$

Form the equation  according to the condition given in the question,

$\[\begin{array}{l}{x^2} + 4x = 21\\ \Rightarrow {x^2} + 4x - 21 = 0\\ \Rightarrow {x^2} + \left( {7 - 3x} \right) - 21 = 0\\ \Rightarrow {x^2} + 7x - 3x - 21 = 0\end{array}\]$

$\[\begin{array}{l} \Rightarrow x\left( {x + 7} \right) - 3\left( {x + 7} \right) = 0\\ \Rightarrow \left( {x - 3} \right)\left( {x + 7} \right) = 0\\ \Rightarrow x - 3 = 0\,\,or\,\,x + 7 = 0\\ \Rightarrow x = 3, - 7\end{array}\]$

Therefore,  The number is either 3 or $-7$.