Question

How do you solve \[14{n^2} + 24n - 5004 = 0\]?

Answer 1

Hint: First compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(i)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.

Complete step by step solution:
First, we have to compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing \[14{n^2} + 24n - 5004 = 0\] with $a{x^2} + bx + c = 0$, we get
$a = 14$, $b = 24$ and $c = - 5004$
Now, we have to substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( {24} \right)^2} - 4\left( {14} \right)\left( { - 5004} \right)$
After simplifying the result, we get
$ \Rightarrow D = 576 + 280224$
$ \Rightarrow D = 280800$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $n = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow n = \dfrac{{ - 24 \pm 60\sqrt {78} }}{{2 \times 14}}$
Divide both numerator and denominator by $4$, we get
$ \Rightarrow n = \dfrac{{ - 6 \pm 15\sqrt {78} }}{7}$
$ \Rightarrow n = - \dfrac{6}{7} \pm \dfrac{{15}}{7}\sqrt {78} $
Final solution: Hence, with the help of formula (i) we obtain the solution to the quadratic equation\[14{n^2} + 24n - 5004 = 0\], which are $n = - \dfrac{6}{7} \pm \dfrac{{15}}{7}\sqrt {78} $.

Note:
We can also find the solution of given quadratic equation by creating a trinomial square on the left side of the equation and using below algebraic identity:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$……(ii)
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$…(iii)
Where, $a$and $b$are any two numbers.
Step by step solution:
First of all, we have to divide both sides of the equation by $2$, we get
$ \Rightarrow 7{n^2} + 12n - 2502 = 0$
Now, multiply both sides of the equation by $7$, we get.
$ \Rightarrow 49{n^2} + 84n - 17514 = 0$
Now, we have to create a trinomial square on the left side of the equation.
$ \Rightarrow {\left( {7n} \right)^2} + 2\left( {7n} \right)\left( 6 \right) + 36 - 17514 = 0$
$ \Rightarrow {\left( {7n} \right)^2} + 2\left( {7n} \right)\left( 6 \right) + {6^2} - \left( {{{15}^2} \cdot 78} \right) = 0$
$ \Rightarrow {\left( {7n + 6} \right)^2} - {\left( {15\sqrt {78} } \right)^2} = 0$
Now, expand this equation using algebraic identity (iii).
$ \Rightarrow \left( {7n + 6 - 15\sqrt {78} } \right)\left( {7n + 6 + 15\sqrt {78} } \right) = 0$
Now, equate both factors to $0$ and find the value of $n$.
So, first simplifying $\left( {7n + 6 - 15\sqrt {78} } \right) = 0$.
Move all constant terms to the right side of the equation.
$ \Rightarrow 7n = - 6 + 15\sqrt {78} $
Divide both sides of the equation by $7$, we get
$ \Rightarrow n = - \dfrac{6}{7} + \dfrac{{15}}{7}\sqrt {78} $
Now, simplifying $\left( {7n + 6 + 15\sqrt {78} } \right) = 0$.
Move all constant terms to the right side of the equation.
$ \Rightarrow 7n = - 6 - 15\sqrt {78} $
Divide both sides of the equation by $7$, we get
$ \Rightarrow n = - \dfrac{6}{7} - \dfrac{{15}}{7}\sqrt {78} $
Final solution: Hence, with the help of formula (ii) and (iii) we obtain the solution to the quadratic equation \[14{n^2} + 24n - 5004 = 0\], which are $n = - \dfrac{6}{7} \pm \dfrac{{15}}{7}\sqrt {78} $.