Question

How do you solve $ \sqrt {2n + 3} = n $ and check your solution ?

Answer 1

Hint:Both the left side and the right side involve the unknown quantity n, so we simplify the equation by squaring both sides. Then the equation obtained is a quadratic polynomial equation. So the given equation is a function of n and can be plotted on the graph by putting different values of n, the values of n at which the value of the function is zero are called the factors of the equation and can be found out by factoring or completing the square method.

Complete step by step answer:
We have $ \sqrt {2n + 3} = n $
On squaring both the sides, we get –
 $
{(\sqrt {2n + 3} )^2} = {n^2} \\
2n + 3 = {n^2} \\
\Rightarrow {n^2} - 2n - 3 = 0 \\
 $

The obtained equation is quadratic and is solved by factorization –
 $
{n^2} - 3n + n - 3 = 0 \\
n(n - 3) + 1(n - 3) = 0 \\
\Rightarrow (n - 3)(n + 1) = 0 \\
\Rightarrow n = 3,\;n = - 1 \\
 $

Now, to check if the obtained is correct or not, we put the obtained values of x in $ \sqrt {2n + 3} $ ,
the values obtained are correct if the answer comes out to be greater than zero –
 $
\sqrt {2(3) + 3} = \sqrt 9 \\
\Rightarrow \sqrt 9 > 0 \\

 $

So, $ n = 3 $ is the correct answer.
 $
\sqrt {2( - 1) + 3} = \sqrt { - 2 + 3} \\
\Rightarrow \sqrt 1 > 0 \\
 $

So, $ n = - 1 $ is the correct answer.

Note:The degree of a polynomial equation is defined as the highest exponent of the polynomial. There exist exactly as many roots in a polynomial equation as its degree.

The roots of an equation are the values of x at which the value of the other variable is zero, that is the roots are simply the x- intercepts. The condition to form factors for factorization of a given equation is that we have to express the coefficient of x that is $ b $ as a sum of two numbers such that their product is equal to the product of $ a $ and $ c $ that is $ a \times c = {b_1} \times {b_2} $