D01-41

Official Solution:

The equation \(x^2 + mx - n = 0\), where \(x\) is a variable and \(m\) and \(n\) are constants, has equal roots. One of the roots of another equation \(y^2 + my + 15 = 0\), where \(y\) is a variable and \(m\) is a constant, is 3. What is the value of \(n\)?

A. \(-\frac{1}{64}\)
B. \(-\frac{1}{16}\)
C. \(-15\)
D. \(-16\)
E. \(-64\)


Since one of the roots of the equation \(y^2 + my + 15 = 0\) is 3, we can substitute \(y = 3\) to get \(3^2+3m+15=0\). Solving for \(m\) gives \(m=-8\).

Next, we substitute \(m=-8\) in the first equation to get \(x^2-8x-n=0\).

Since \(x^2-8x-n=0\) has equal roots, it can be expressed as the square of the difference \((x - k)^2 = x^2-2xk+k^2\). Using this form, we can see that \(-2xk=-8x\), and solving for \(k\), we get \(k=4\). Therefore, we can substitute \(k\) into the equation \((x - k)^2\) to get \((x - 4)^2\). Equating this to \(x^2-8x-n=0\), we have \((x - 4)^2=x^2-8x-n\). Expanding this expression gives us \(x^2-8x+16=x^2-8x-n\). Simplifying, we get \(n=-16\).

Alternatively, if you are familiar with the concept of the discriminant, you can recall that the equation \(x^2-8x-n=0\) having equal roots means that its discriminant must be zero. The discriminant of a quadratic equation of the form \(ax^2+bx+c=0\) is given by the expression \(b^2-4ac\). In this case, we have \(a=1\), \(b=-8\), and \(c=-n\). Therefore, the discriminant is \(b^2-4ac=(-8)^2-4(1)(-n)=64+4n\). Since the discriminant must be zero, we have \(64+4n=0\), which gives \(n=-16\).


Answer: D