
Solve the following equation: ${{n}^{2}}+301n-12132=0$.
Answer
614.4k+ views
Hint: In order to solve this question, we should know that for any quadratic equation like, $a{{x}^{2}}+bx+c=0$, we can use Shridharacharya’s formula to find roots, that is, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Also, we should be very careful while doing the calculations.
Complete step-by-step solution -
In this question, we have been asked to solve a quadratic equation, which is ${{n}^{2}}+301n-12132=0$. To solve this question, we should know the concept of Shridharacharya’s formula for quadratic equations, which states that for a quadratic equation, $a{{x}^{2}}+bx+c=0$, the roots are, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, for x = n, a = 1, b = 301 and c = -12132, we can write the roots of the given equation, ${{n}^{2}}+301n-12132=0$ as,
$n=\dfrac{-\left( 301 \right)\pm \sqrt{{{\left( 301 \right)}^{2}}-4\left( 1 \right)\left( -12132 \right)}}{2\left( 1 \right)}$
Now, we know that ${{\left( 301 \right)}^{2}}=90601$ and $\left( 4 \right)\times \left( 12132 \right)=48528$. So, we will write them in the equation and get the roots as,
$n=\dfrac{-\left( 301 \right)\pm \sqrt{90601+48528}}{2}$
We can further simplify and write it as,
$n=\dfrac{-301\pm \sqrt{139129}}{2}$
Now, we know that $\sqrt{139129}=373$. So, we will put this value and get the roots as follows,
$n=\dfrac{-301\pm 373}{2}$
We can further write it as,
$n=\dfrac{-301+373}{2}$ and $n=\dfrac{-301-373}{2}$
Which is the same as follows,
$n=\dfrac{72}{2}$ and $n=\dfrac{-674}{2}$
$\Rightarrow n=36$ and $n=-337$.
Hence, we have obtained the value of the roots of the given quadratic equation, ${{n}^{2}}+301n-12132=0$ as, n = 36 and -337.
Note: While solving this question, there is a high chance of making calculation mistakes, so to verify our answer, we can use the formula of the sum of roots, that is sum of roots $=\dfrac{-b}{a}$ and the formula for the product of roots $=\dfrac{c}{a}$. We can also solve this question by using the middle term splitting method, but since the middle term in this question is 301, and the constant is also 12132, it will be difficult to apply that method for such big numbers.
Complete step-by-step solution -
In this question, we have been asked to solve a quadratic equation, which is ${{n}^{2}}+301n-12132=0$. To solve this question, we should know the concept of Shridharacharya’s formula for quadratic equations, which states that for a quadratic equation, $a{{x}^{2}}+bx+c=0$, the roots are, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, for x = n, a = 1, b = 301 and c = -12132, we can write the roots of the given equation, ${{n}^{2}}+301n-12132=0$ as,
$n=\dfrac{-\left( 301 \right)\pm \sqrt{{{\left( 301 \right)}^{2}}-4\left( 1 \right)\left( -12132 \right)}}{2\left( 1 \right)}$
Now, we know that ${{\left( 301 \right)}^{2}}=90601$ and $\left( 4 \right)\times \left( 12132 \right)=48528$. So, we will write them in the equation and get the roots as,
$n=\dfrac{-\left( 301 \right)\pm \sqrt{90601+48528}}{2}$
We can further simplify and write it as,
$n=\dfrac{-301\pm \sqrt{139129}}{2}$
Now, we know that $\sqrt{139129}=373$. So, we will put this value and get the roots as follows,
$n=\dfrac{-301\pm 373}{2}$
We can further write it as,
$n=\dfrac{-301+373}{2}$ and $n=\dfrac{-301-373}{2}$
Which is the same as follows,
$n=\dfrac{72}{2}$ and $n=\dfrac{-674}{2}$
$\Rightarrow n=36$ and $n=-337$.
Hence, we have obtained the value of the roots of the given quadratic equation, ${{n}^{2}}+301n-12132=0$ as, n = 36 and -337.
Note: While solving this question, there is a high chance of making calculation mistakes, so to verify our answer, we can use the formula of the sum of roots, that is sum of roots $=\dfrac{-b}{a}$ and the formula for the product of roots $=\dfrac{c}{a}$. We can also solve this question by using the middle term splitting method, but since the middle term in this question is 301, and the constant is also 12132, it will be difficult to apply that method for such big numbers.
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