DaalChawal
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Are You Struggling with Quadratic Equations?
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Hi DaalChawal,
We can write
$$(x-\alpha)(x-\beta)+(x-\gamma)(x-\delta)=2x^2-(\alpha+\beta+\gamma+\delta)x+(\alpha\beta+\gamma\delta)=0\implies $$
$$x^2-\frac 12(\alpha+\beta+\gamma+\delta)x+\frac 12(\alpha\beta+\gamma\delta) = 0\tag 1$$
Since $m$ and $n$ are roots, we must have
$$(x-m)(x-n)=0\implies x^2-(m+n)x+mn=0\tag 2$$
The coefficients of equations (1) and (2) must be the same, so we have:
$$\begin{cases}m+n=\frac 12(\alpha+\beta+\gamma+\delta)\\mn=\frac 12(\alpha\beta+\gamma\delta)\end{cases}\tag 3$$
We have
$$2(x-m)(x-n)-(x-\alpha)(x-\beta)=x^2-(2(m+n)-\alpha-\beta)x+(2mn-\alpha\beta)=0$$
Let its roots be $x_1$ and $x_2$, then we must have:
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
Again, the coefficients must match, so we have
$$\begin{cases}x_1+x_2=2(m+n)-\alpha-\beta \\ x_1x_2=2mn-\alpha\beta\end{cases}\tag 4$$
Substitute (3) in (4) to find:
$$\begin{cases}x_1+x_2=2\cdot\frac 12(\alpha+\beta+\gamma+\delta)-\alpha-\beta = \gamma+\delta \\
x_1x_2=2\cdot \frac 12(\alpha\beta+\gamma\delta)-\alpha\beta = \gamma\delta\end{cases}\tag 5$$
Therefore (B) is the correct answer.
We can write
$$(x-\alpha)(x-\beta)+(x-\gamma)(x-\delta)=2x^2-(\alpha+\beta+\gamma+\delta)x+(\alpha\beta+\gamma\delta)=0\implies $$
$$x^2-\frac 12(\alpha+\beta+\gamma+\delta)x+\frac 12(\alpha\beta+\gamma\delta) = 0\tag 1$$
Since $m$ and $n$ are roots, we must have
$$(x-m)(x-n)=0\implies x^2-(m+n)x+mn=0\tag 2$$
The coefficients of equations (1) and (2) must be the same, so we have:
$$\begin{cases}m+n=\frac 12(\alpha+\beta+\gamma+\delta)\\mn=\frac 12(\alpha\beta+\gamma\delta)\end{cases}\tag 3$$
We have
$$2(x-m)(x-n)-(x-\alpha)(x-\beta)=x^2-(2(m+n)-\alpha-\beta)x+(2mn-\alpha\beta)=0$$
Let its roots be $x_1$ and $x_2$, then we must have:
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
Again, the coefficients must match, so we have
$$\begin{cases}x_1+x_2=2(m+n)-\alpha-\beta \\ x_1x_2=2mn-\alpha\beta\end{cases}\tag 4$$
Substitute (3) in (4) to find:
$$\begin{cases}x_1+x_2=2\cdot\frac 12(\alpha+\beta+\gamma+\delta)-\alpha-\beta = \gamma+\delta \\
x_1x_2=2\cdot \frac 12(\alpha\beta+\gamma\delta)-\alpha\beta = \gamma\delta\end{cases}\tag 5$$
Therefore (B) is the correct answer.
DaalChawal
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Thank you Sir