If-the-equations-x-2-3x-a-0-and-x-2-ax-3-0-have-a-common-root-then-a-possible-value-of-a-is-A-3-B-1-C-2-D-2-

Question Number 140974 by EnterUsername last updated on 14/May/21

If the equations x^{2} - 3 x + a = 0 and x^{2} + a x - 3 = 0

have a common root, then a possible value of a is

(A) 3 (B) 1 (C) - 2 (D) 2

Commented by mohammad17 last updated on 14/May/21

x^{2} + a x - 3 = 0 \to x + a - \frac{3}{x} = 0 \to (1)

x^{2} - 3 x + a = 0 \to (2)

f r o m (2) - (1)

x^{2} - 4 x + \frac{3}{x} = 0 \Rightarrow x^{3} - 4 x^{2} + 3 = 0

x = 1, x = \frac{3 - \sqrt{21}}{2}, x = \frac{3 + \sqrt{21}}{2}

s i n c e : x = 1 \Rightarrow a = 2

s i n c e : x = \frac{3 + \sqrt{21}}{2} \Rightarrow a = - 3 c a n c e l

s i n c e x = \frac{3 - \sqrt{21}}{2} \Rightarrow a = - 3 c a n c e

\Rightarrow a = 2

Commented by EnterUsername last updated on 14/May/21

Thanks you!

Commented by mohammad17 last updated on 14/May/21

y o u a r e w e l c o m e

Answered by henderson last updated on 14/May/21

the answer is a = 2 .

Answered by physicstutes last updated on 14/May/21

I will say from experience a = 2, but if you need a proper

working step, you could say

let the first equation have roots α and β

and let the second equation have root β and ϑ

notice that β is the common root .

α + β = 3 \dots . . (i) and α β = a \dots . . (i i)

β + ϑ = - a \dots . . (i i) and β ϑ = - 3 \dots . . (i v)

from (i v) ϑ = - \frac{3}{β}

\Rightarrow β - \frac{3}{β} = - a

also α = \frac{a}{β}

\Rightarrow \frac{a}{β} + β = 3 and \frac{3}{β} - β = a

a + β^{2} = 3 β

and 3 - β^{2} = a β

add them and get : 3 + a = β (3 + a) \Rightarrow β = 1

now 1 - \frac{3}{1} = - a \Rightarrow a = 2

Commented by EnterUsername last updated on 14/May/21

Thanks for the idea!