# When 18 is subtracted from the square of a number, the result is 3 times the number. Find the...

## Question:

When 18 is subtracted from the square of a number, the result is 3 times the number. Find the negative solution.

Quadratic Equations: A mathematical equation is called a quadratic if it has degree 2 means the highest possible power of a variable is 2 in the equation. It consists of real constants and variables.

{eq}\displaystyle px^{2}+qx+r = 0 \text{ where } p \neq 0 {/eq}

Here,

• {eq}\displaystyle{ p }{/eq}, {eq}\displaystyle{ q }{/eq} and {eq}\displaystyle{ r }{/eq} are the real constants
• {eq}\displaystyle{ x }{/eq} is a variable

A quadratic equation has two solutions that can be real or complex depending on the value of their discriminant.

{eq}\displaystyle D = q^{2}-4pr {/eq}

Here, {eq}\displaystyle{ D }{/eq} is the discriminant.

Factorization method:

In Factorization, we can write a quadratic equation as the multiplication of two linear equations or two factors.

{eq}\displaystyle px^{2}+qx+r = 0 {/eq}

{eq}\displaystyle p(x-\alpha)(x-\beta) = 0 {/eq}

Here,

{eq}\displaystyle x-\alpha = 0 {/eq} or {eq}\displaystyle x-\beta = 0 {/eq}

So, {eq}\displaystyle x = \alpha {/eq} or {eq}\displaystyle x = \beta {/eq} are the solutions of the quadratic equation.

{eq}\displaystyle{ \\ }{/eq}

Let {eq}\displaystyle{ n }{/eq} be a number.

Given that when 18 is subtracted from the square of a number, the result is 3 times the number.

So the corresponding equation is:

{eq}\displaystyle n^{2}-18 = 3n {/eq}

Now transfer all the terms on one side of the equation.

{eq}\displaystyle n^{2}-3n-18 = 0 {/eq}

For factorization, we have to break the middle term {eq}(-3n) {/eq} into two terms such that the multiplication of their coefficients is equal to the multiplication of coefficient of {eq}n^{2} {/eq} and the constant term {eq}(-18). {/eq}

So let us check if the following satisfies the condition:

{eq}\displaystyle -3n = 6n-3n {/eq}

{eq}\displaystyle (6 \times (-3)) = 1 \times (-18) {/eq}

This satisfies the condition.

Now factorize the equation:

{eq}\displaystyle n^{2}-6n+3n-18 = 0 {/eq}

Now take common:

{eq}\displaystyle n(n-6)+3(n-6) = 0 {/eq}

{eq}\displaystyle (n-6)(n+3) = 0 {/eq}

So, {eq}\displaystyle n= 6 {/eq} or {eq}\displaystyle n = -3 {/eq}

So {eq}\displaystyle \color{blue}{n = -3} {/eq} is the negative solution of the equation.