Question

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One of the roots of a quadratic equation a²x² - 3abx + 2b² = 0.
(i) -2a/b
(ii) 2a/b
(iii) 2b/a
(iv) -2b/a

[ Need perfect answer, please ]

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Answer 1

Answer:

a2x2−3abx+2b2=0

=>a2x2−abx−2abx+2b2=0

=>ax(ax−b)−2b(ax−b)=0

=>(ax−b)(ax−2b)=0

=>ax−b=0orax−2b=0

=>x=a/b or x= 2b/a

(iii) 2b/a

Answer 2

Answer:

a²x²-3abx+2b²=0

or, {(ax)²-2.ax.3b/2+(3b/2)²}+2b²-9b²/4=0

or, (ax-3b/2)²+(8b²-9b²)/4=0

or, (ax-3b/2)²-(b/2)²=0

or, (ax-3b/2+b/2)(ax-3b/2-b/2)=0

or, (ax-b)(ax-2b)=0

Either, ax-b=0

or, ax=b

or, x=b/a

Or, ax-2b=0

or,  ax=2b

or, x=2b/a

∴, the roots of the given equations are: b/a, 2b/a. Ans.

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