Question

The product of two consecutive positive integers is $120$. How do you find the integers?

Answer 1

We will start off by explaining the term “more than the other term”. Then we will be mentioning the steps to solve such types of questions. We have considered an unknown variable as $w$ to further simplify and evaluate the terms. Also, we will mention some other rules as well.

Complete step by step answer:
We will start off by considering $x$ as the first number, so that the other term will be $x + 1$. So according to the given condition, we can write,
$x \times (x + 1) = 120$
Now if we open the brackets, we will get a quadratic.
${x^2} + x = 120 \\\ {x^2} + x - 120 = 0 \\\$
We will start off by reducing any reducible terms in the equation if possible.
${x^2} + x - 120 = 0$
Now we will factorise the terms in the equation.
${x^2} + x - 120 = 0$
Now we will try to factorise by using the quadratic formula which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
To substitute the values, we first compare and evaluate the values from the general form of
quadratic equation. The general form of the quadratic equation is given by $a{x^2} + bx + c = 0$.
When we compare the terms, we get the values as,
$a = 1 \\\ b = 1 \\\ c = - 120 \\\$
Now substitute all these values in the quadratic formula, to evaluate the value of the variable.

$$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ x = \dfrac{{ - (1) \pm \sqrt {{{(1)}^2} - 4(1)( - 120)} }}{{2(1)}} \\\ x = \dfrac{{ - (1) \pm \sqrt {(1) + 4(1)(120)} }}{{2(1)}} \\\ x = \dfrac{{ - 1 \pm \sqrt {(1) + 480} }}{2} \\\ x = \dfrac{{ - 1 \pm \sqrt {481} }}{2} \\\$$

Now, we know that $\sqrt {481}$ is not a perfect square, which means this discriminant will not give us rational solutions and according to the given condition, the numbers are positive integers.
Hence, there are no such positive integers.
Additional Information:
Expression is a mathematical sentence which has numbers, variables, and operations and there is no equal sign. Simplify means to break down an expression to its simplest forms. Variable is a number that we don’t know, or which can change. Algebraic expressions are useful because they represent the value of an expression for all of the values a variable can take on. Sometimes in math, we describe an expression with a phrase. When we describe an expression in words that includes a variable, we are describing an algebraic expression, an expression with a variable.

Note: While converting orders do not matter for addition and multiplication. But order is important for subtraction and division. Make sure that you read the statement twice before translating it to an expression. Pay extra attention to the statements where multiplication and division is involved.