Question

How do you solve \[{x^2} - 3x = 40\] ?

Answer 1

Hint: In this question, the unknown variables of the algebraic expression are raised to some non-negative integer as a power so it is a polynomial equation. The highest exponent is 2 so the given equation is a polynomial equation of degree 2, hence it is a quadratic equation and has exactly two solutions. We have to solve this equation, that is, we have to find its solutions. Solutions of an equation are defined as the values of the x for which the given function has a value zero or when we plot this function on the graph, we see that the points on which the y-coordinate is zero are known as the solutions of the equation, thus they are simply the x-intercepts.

Complete step-by-step solution:
We are given that
\[{x^2} - 3x = 40\]
Now we will take all the terms to one side such that they are equal to zero to convert it into standard form –
\[{x^2} - 3x - 40 = 0\]
The obtained equation is in standard form, on factorization we get –
$
  {x^2} - 3x - 40 = 0 \\
   \Rightarrow {x^2} - 8x + 5x - 40 = 0 \\
   \Rightarrow x(x - 8) + 5(x - 8) = 0 \\
   \Rightarrow (x - 8)(x + 5) = 0 \\
   \Rightarrow x - 8 = 0,\,x + 5 = 0 \\
   \Rightarrow x = 8,\,x = - 5 \\
 $
Hence when \[{x^2} - 3x - 40 = 0\] , we get $x = 8$ and $x = - 5$ .

Note: $a{x^2} + bx + c = 0$ is the standard form of a quadratic equation. To find the factors of the given equation, we compare the given equation and the standard equation and get the values of a, b and c. Then we will try to write b as a sum of two numbers such that their product is equal to the product of a and c, that is, ${b_1} \times {b_2} = a \times c$ , this method is known as factorization. We find the value of ${b_1}$ and ${b_2}$ by hit and trial. We move to some other methods like quadratic formula, graphing, and completing the square method, if we are not able to solve an equation by factorization.