Thread: Two Quadratic Equations With (a) Common Root, and an Unknown Coefficient

1. Two Quadratic Equations With (a) Common Root, and an Unknown Coefficient

<B>The question is:</B>
If x<sup>2</sup> + 4ax + 3 = 0 and 2x<sup>2</sup> + 3ax – 9 = 0 have a common root, then the value of ‘a’ is?

(A) +3,-3
(B) +1,-1
(C) Only 1
(D) +2,-2

<B>My approach</B>
At first, I simply plugged in every option and figured out the answer, but, I want to know if there is any way I could have done this had I not been given options?

I started out by considering 'b' as the common root, substituting it for 'x' in both the equations, and then equating the equations. Then...I got stuck...and I've been stuck ever since...

Any help would be really appreciated.
Thanks!

2. Note that $\displaystyle (x- x_1)(x- x_2)= x^2- (x_1+ x_2)x+ x_1x_2$. That is, that in the general form $\displaystyle x^2- bx+ c= 0$, b is the sum of the roots and c is their product.

To get the second equation, $\displaystyle 2x^2+ 3ax- 9= 0$ into that form I would first divide through by 2: $\displaystyle x^2- (3/2)x- 9/2$ with b= -3a/2, c= -9/2. The first equation, $\displaystyle x^2+ 4ax+ 3= 0$ has b= -4a and c= 3.

If we call the common root, "B" and the other roots "A" and "C", we must have A+ B= -3a/2, B+ C= -4a, AB= -9/2, and BC= 3. That gives four equations to solve for the four unknowns, A, B, C, and a. From AB= -9/2, for example, A= -9/(2B) and from BC= 3, C= 3/B. Putting those into A+ B= -3a/2, and B+ C= -4a, -9/(2B)+ B= -3a/2 and B+ 3/B= -4a so you are now reduced to two equations in B and a.

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common root in one quadratic and one linear

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