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How do you solve for $n$ in $\left( {9n + 4} \right) + \left( {3{n^2} - 0.75} \right) = 180$?

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Last updated date: 21st Jul 2024
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Views today: 8.82k
Answer
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Hint:The general form of a quadratic equation is \[a{x^2} + bx + c = 0\], where $a \ne 0$. The values of $x$ which satisfy the equation are called the roots or the solutions. These roots can be real or imaginary. There are two ways to find solutions/roots of $x$– comparing method and discriminant method. Let us solve the given equation for the value of $n$ by quadratic formula.

Complete step by step solution:
Before solving this question, we will first have to convert the given equation into general form, then compare the given equation with the standard quadratic equation, which is$a{x^2} + bx + c = 0$,
where$a \ne 0$. After comparing both the equations with each other, we will have to find the values of $a, b$ and $c$. Next, we will have to substitute the obtained values in the quadratic formula.
Given is$\left( {9n + 4} \right) + \left( {3{n^2} - 0.75} \right) = 180$
Simplifying the equation,
$ \Rightarrow 3{n^2} + 9n + 4 - 0.75 - 180 = 0$
Since, $\left( {0.75 = \dfrac{{75}}{{100}} = \dfrac{3}{4}} \right)$
$ \Rightarrow 3{n^2} + 9n + 4 - \dfrac{3}{4} - 180 = 0$
Solving the constants, we get,
$4 - \dfrac{3}{4} - 180 = \dfrac{{16 - 3 - 360}}{4} = - \dfrac{{707}}{4}$
Now, we have
$ \Rightarrow 3{n^2} + 9n - \dfrac{{707}}{4} = 0$
Divide the entire equation by$3$and we get,
$ \Rightarrow {n^2} + 3n - \dfrac{{707}}{{12}} = 0$
Next, we use the quadratic formula. We know that quadratic formula is: $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using the same we get,
$
\Rightarrow n = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 1 \right)\left( { - \dfrac{{707}}{{12}}}
\right)} }}{{2\left( 3 \right)}} \\
\Rightarrow n = - \dfrac{3}{2} \pm \sqrt {\dfrac{{367}}{6}} \\
\Rightarrow n = - \dfrac{3}{2} \pm \dfrac{{\sqrt {2202} }}{6} \\
$
So, the two values of $n$ we get are,
$
{n_1} = \dfrac{{\sqrt {2202} - 9}}{6} \\

{n_2} = - \dfrac{1}{6}\left( {9 + \sqrt {2202} } \right) \\
$
Therefore, the values of $n$ are $\dfrac{{\sqrt {2202} - 9}}{6}$ and $ - \dfrac{1}{6}\left( {9 + \sqrt {2202} } \right)$.

Note: In the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, the expression $\sqrt {{b^2} - 4ac} $ is called discriminant and is often denoted by $D$. Now, depending upon the quadratic equation, the value of discriminant changes and therefore, the value of roots also change. If $D$ is positive or greater than zero, then the two roots of the equation are real. If $D$ is zero, then roots are real but if $D$ is negative or less than zero, then roots are not real.