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if there is a common root of the equation ax²+a²x+1=0 and bx²+b²x+1=0 show that their other roots satisfy the equation abx²+x+a²b²=0​

User Jendrusk
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Answer: Yes!

Explanation:

Answer: Yes, the other roots of the equations ax2+a2x+1=0 and bx2+b2x+1=0 will satisfy the equation abx2+x+a2b2=0.

To show this, let us assume that x1 and x2 are the other two roots of the equations ax2+a2x+1=0 and bx2+b2x+1=0 respectively. Then, substituting x1 and x2 in the equation abx2+x+a2b2=0, we get:

abx12 + x1 + a2b2 = abx22 + x2 + a2b2

Rearranging the terms, we get:

(abx12 - abx22) + (x1 - x2) = 0

Now, since x1 and x2 are the roots of the two equations, we can write:

a(x1 - x2) (bx1 - bx2) = 0

Therefore,

abx12 - abx22 + x1 - x2 = 0

Hence, we can conclude that the other roots of the equations ax2+a2x+1=0 and bx2+b2x+1=0 will satisfy the equation abx2+x+a2b2=0.

User Scott Chapman
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