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By completing square method find the roots of the following quadratic equation
$2{n^2} - 7n + 3 = 0$.

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint: Here, we will simply add and subtract square of a constant and write the term containing $n$ such that it forms a term in a form $2ab$, thus, this will help us to apply the square identities and thus, ‘complete the square’ and solve it further to find the required roots of the given quadratic equation.

Formula Used:
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

Complete step-by-step answer:
The given quadratic equation is:
$2{n^2} - 7n + 3 = 0$
Now, dividing both sides by 2, we get,
${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$
Now, since, we are required to solve this question using completing the square, hence, we will write this quadratic equation as:
${\left( n \right)^2} - 2\left( n \right)\left( {\dfrac{7}{4}} \right) + {\left( {\dfrac{7}{4}} \right)^2} - {\left( {\dfrac{7}{4}} \right)^2} + \dfrac{3}{2} = 0$
Hence, even if this quadratic equation was not a perfect square, but we tried to make it a perfect square by adding and subtracting the square of a constant such that, we complete the square by making the identity, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Hence, using this identity, we get,
${\left( {n - \dfrac{7}{4}} \right)^2} - \dfrac{{49}}{{16}} + \dfrac{3}{2} = 0$
$ \Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} + \dfrac{{ - 49 + 24}}{{16}} = 0$
Hence, we get,
$ \Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} - \dfrac{{25}}{{16}} = 0$
Adding $\dfrac{{25}}{{16}}$ on both sides,
$ \Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} = \dfrac{{25}}{{16}}$
$ \Rightarrow {\left( {n - \dfrac{7}{4}} \right)^2} = {\left( {\dfrac{5}{4}} \right)^2}$
Taking square root on both sides, we get
$ \Rightarrow \left( {n - \dfrac{7}{4}} \right) = \pm \dfrac{5}{4}$
Adding $\dfrac{7}{4}$ on both the sides, we get,
$ \Rightarrow n = \pm \dfrac{5}{4} + \dfrac{7}{4}$
Hence,
$n = \dfrac{5}{4} + \dfrac{7}{4} = \dfrac{{5 + 7}}{4} = \dfrac{{12}}{4} = 3$
Or $n = - \dfrac{5}{4} + \dfrac{7}{4} = \dfrac{{7 - 5}}{4} = \dfrac{2}{4} = \dfrac{1}{2}$
Therefore, the roots of the given quadratic equation $2{n^2} - 7n + 3 = 0$ are 3 and $\dfrac{1}{2}$
Thus, this is the required answer.

Note:
If in the question, it was not mentioned that we have to use the method of completing the square, then, we could have used the quadratic formula to solve the given quadratic equation.
Given quadratic equation is $2{n^2} - 7n + 3 = 0$
Now, dividing both sides by 2, we get,
${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$
Comparing this with the general quadratic equation i.e. $a{x^2} + bx + c = 0$
We have,
$a = 1$, $b = \dfrac{{ - 7}}{2}$ and $c = \dfrac{3}{2}$
Now, we can find the roots of a quadratic equation using the quadratic formula, $n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Hence, for the equation ${n^2} - \dfrac{7}{2}n + \dfrac{3}{2} = 0$, substituting$a = 1$, $b = \dfrac{{ - 7}}{2}$and $c = \dfrac{3}{2}$, we get
$n = \dfrac{{\dfrac{7}{2} \pm \sqrt {{{\left( {\dfrac{{ - 7}}{2}} \right)}^2} - 4\left( 1 \right)\left( {\dfrac{3}{2}} \right)} }}{{2\left( 1 \right)}}$
$ \Rightarrow n = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{49}}{4} - 6} }}{2} = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{49 - 24}}{4}} }}{2} = \dfrac{{\dfrac{7}{2} \pm \sqrt {\dfrac{{25}}{4}} }}{2}$
Solving further, we get,
\[ \Rightarrow n = \dfrac{{\dfrac{7}{2} \pm \dfrac{5}{2}}}{2} = \dfrac{{7 \pm 5}}{4}\]
Hence, we get,
\[n = \dfrac{{7 + 5}}{4} = \dfrac{{12}}{4} = 3\]
Or \[n = \dfrac{{7 - 5}}{4} = \dfrac{2}{4} = \dfrac{1}{2}\]
Therefore, the roots of the given quadratic equation $2{n^2} - 7n + 3 = 0$ are 3 and $\dfrac{1}{2}$
Thus, this is the required answer.
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