Please answer fast.tommorow is board exam
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GIVEN
ax^2 + a = a^2x + x
ON ONE SIDE
ax^2 + a - a^2x - x = 0
REARRANGE
ax^2 - x - a^2x + a = 0
x (ax - 1) - a (ax - 1) = 0
( ax - 1) (x - a) = 0
( ax - 1) = 0 & (x - a) = 0
ax = 1 & x = a
x = 1 / a & x = a
Therefore, the roots of said equation are
1 / a & a.
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Answer:
ax² + a = a²x + x
ax² - a²x - x + a = 0
ax(x - a) - 1(x - a) = 0
(ax - 1) × (x - a) = 0
x = 1/a , a → roots of the equation
hope its helpful....................
Really thanks for the answer
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