Question

If $x^{2}-ax-21=0$ and $x^{2}-3ax+35=0$ with $a>0$ have a common root, then $a$ equals:

• A. 1
• B. 2
• C. 4
• D. 5

Hint:

For two quadratics with a common root, subtract to eliminate $x^2$ and solve for the root in terms of parameters; then substitute back.

Answer 1

The Correct Option is C

Let the common root be $r$. Subtract the equations: \[ (-3a\,r+35)-(-a\,r-21)=0 \;\Rightarrow\; -2a r+56=0 \;\Rightarrow\; r=\frac{28}{a}. \] Plug into $r^2-ar-21=0$: \[ \left(\frac{28}{a}\right)^2-a\left(\frac{28}{a}\right)-21=0 \;\Rightarrow\; \frac{784}{a^2}-49=0 \;\Rightarrow\; a^2=16 \;\Rightarrow\; a=4 \;(a>0). \] 

Answer 2

Instead of eliminating the squared term directly, we can use the standard cross-multiplication rule for two quadratics sharing a common root, and then check which option is consistent with it.

1. Option A (\(a=1\)): Substituting \(a=1\) does not satisfy the relation obtained below, so it is rejected.
2. Option B (\(a=2\)): This also fails the relation derived below.
3. Option C (\(a=4\)): Writing the two equations as \(x^2-ax-21=0\) and \(x^2-3ax+35=0\), the common root \(r\) satisfies \[ r=\frac{(-a)(35)-(-3a)(-21)}{(-21)(1)-(35)(1)}=\frac{-98a}{-56}=\frac{7a}{4}, \] and also \[ r=\frac{(-21)(1)-(35)(1)}{(1)(-3a)-(1)(-a)}=\frac{-56}{-2a}=\frac{28}{a}. \] Equating the two expressions for \(r\): \[ \frac{7a}{4}=\frac{28}{a}\;\Rightarrow\;7a^2=112\;\Rightarrow\;a^2=16\;\Rightarrow\;a=4\ (\text{since } a>0). \] This matches option C exactly.
4. Option D (\(a=5\)): This does not satisfy \(a^2=16\), so it is rejected.

Only \(a=4\) satisfies the cross-multiplication condition for a common root, confirming option C.

Hence, the correct answer is option C: \(4\).