If x² + 4ax + 3 = 0 and 2x² + 3ax - 9 = 0 have a common root , then the value of "a" is what ??

Answer :

TPS
Let the common root be m. Put in the equations;
m² + 4am + 3 = 0   --------------(1)
2m² + 3am - 9 = 0   --------------(2)
multiply equation (1) by 2
2m² + 8am + 6 = 0   -------------(3)
(note that equation 1 and 3 are the same)

subtract (2) from (3); you will get
(2m² + 8am + 6) - (2m² + 3am - 9) = 0
⇒ 5am + 15 = 0
⇒ 5am = -15
⇒ m = [tex]- \frac{15}{5a} =- \frac{3}{a} [/tex]

from equation (1),

[tex] (- \frac{3}{a})^{2} + 4a(- \frac{3}{a})+3=0\\ \\ \frac{9}{a^2} - 12+3=0\\ \\ \frac{9}{a^2} -9=0\\ \\\frac{9}{a^2}=9\\ \\a^2= \frac{9}{9}=1\\ \\a=+1\ or\ -1 [/tex]

So values of a are +1 and -1.


Answer:

Answer is +-1 and +-3

Step-by-step explanation:

Hope the answer will help you

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