
If the roots of the equation ${{x}^{2}}+2mx+{{m}^{2}}-2m+6=0$ are the same, then the value of $m$ will be
A. $3$
B. $0$
C. $2$
D. $-1$
Answer
164.1k+ views
Hint: In this question, we are to find the value of $m$ in the given quadratic equation. Since the question states that the roots are the same, we must first assume that there are two identical roots for this equation. From these roots, we will then find the sum and product of the roots of the given equation, and by simplifying the equations of the sum and product of roots, we will obtain the value of m that we need.
Formula UsedConsider a quadratic equation $a{{x}^{2}}+bx+c=0$.
Then, its roots are $\alpha $ and $\beta $. So, the sum of the roots and the product of the roots is formulated as:
$\alpha +\beta =\dfrac{-b}{a};\alpha \beta =\dfrac{c}{a}$
Complete step by step solution:The given quadratic equation is ${{x}^{2}}+2mx+{{m}^{2}}-2m+6=0$ and its roots are $\alpha $ and $\beta $.
So, comparing the given equation with the standard equation, we get
$a=1;b=2m;c={{m}^{2}}-2m+6$
Then, the sum of the roots of the given equation is obtained as
$\alpha +\beta =\dfrac{-2m}{1}=-2m\text{ }...(1)$
And the product of the roots of the given equation is
$\alpha \beta =\dfrac{{{m}^{2}}-2m+6}{1}={{m}^{2}}-2m+6\text{ }...(2)$
It is given that the roots are the same for the given equation. So, we can write $\alpha =\beta $. Then, substituting this value in expression (1), we get
$\begin{align}
& \alpha +\alpha =-2m \\
& \Rightarrow 2\alpha =-2m \\
& \Rightarrow \alpha =-m \\
\end{align}$
On substituting the obtained value in (2), we get
$[\begin{align}
& \alpha \times \alpha ={{m}^{2}}-2m+6 \\
& \Rightarrow {{\alpha }^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow {{(-m)}^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow {{m}^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow 2m=6 \\
& \therefore m=3 \\
\end{align}]$
Thus, this is the required value.
Option ‘A’ is correct
Note: Here, we need to remember the sum and product formulae. From the given relation i.e., equal roots for the given equation we can find the required value. And be sure with the signs in the formulae. Otherwise, the entire calculation may go in the wrong way.
Formula UsedConsider a quadratic equation $a{{x}^{2}}+bx+c=0$.
Then, its roots are $\alpha $ and $\beta $. So, the sum of the roots and the product of the roots is formulated as:
$\alpha +\beta =\dfrac{-b}{a};\alpha \beta =\dfrac{c}{a}$
Complete step by step solution:The given quadratic equation is ${{x}^{2}}+2mx+{{m}^{2}}-2m+6=0$ and its roots are $\alpha $ and $\beta $.
So, comparing the given equation with the standard equation, we get
$a=1;b=2m;c={{m}^{2}}-2m+6$
Then, the sum of the roots of the given equation is obtained as
$\alpha +\beta =\dfrac{-2m}{1}=-2m\text{ }...(1)$
And the product of the roots of the given equation is
$\alpha \beta =\dfrac{{{m}^{2}}-2m+6}{1}={{m}^{2}}-2m+6\text{ }...(2)$
It is given that the roots are the same for the given equation. So, we can write $\alpha =\beta $. Then, substituting this value in expression (1), we get
$\begin{align}
& \alpha +\alpha =-2m \\
& \Rightarrow 2\alpha =-2m \\
& \Rightarrow \alpha =-m \\
\end{align}$
On substituting the obtained value in (2), we get
$[\begin{align}
& \alpha \times \alpha ={{m}^{2}}-2m+6 \\
& \Rightarrow {{\alpha }^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow {{(-m)}^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow {{m}^{2}}={{m}^{2}}-2m+6 \\
& \Rightarrow 2m=6 \\
& \therefore m=3 \\
\end{align}]$
Thus, this is the required value.
Option ‘A’ is correct
Note: Here, we need to remember the sum and product formulae. From the given relation i.e., equal roots for the given equation we can find the required value. And be sure with the signs in the formulae. Otherwise, the entire calculation may go in the wrong way.
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