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If the roots of the equation \(x^2 + mx + n = 0\) are increased by the same quantity \(k\), then they become the roots of the equation \(x^2 + nx + m = 0\). What is the value of \((m+n)\)?

This question was previously asked in
CDS 2 2026 Maths Question Paper (13-Sep-2026)
The correct answer is

-4

Let the roots of \(x^2+mx+n=0\) be \(p\) and \(q\), so \(p+q=-m\) and \(pq=n\).

The new roots \(p+k\) and \(q+k\) satisfy \(x^2+nx+m=0\), so \((p+k)+(q+k)=-n\) and \((p+k)(q+k)=m\).

From the sum: \(-m+2k=-n \Rightarrow 2k=m-n\).

From the product: \(n-km+k^2=m\). Substituting \(k=\frac{m-n}{2}\) and simplifying gives \((n-m)(n+m+4)=0\).

Since \(k\) is a genuine (non-zero) shift, \(n \neq m\), so \(m+n=-4\).

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