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Factorise: 4 a 2 + 1 4 a 2 − 2 − 6 a + 3 2 a - Mathematics

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प्रश्न

Factorise:

`4"a"^2 + (1)/(4"a"^2) - 2 - 6"a" + (3)/(2"a")`

योग
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उत्तर

`4"a"^2 + (1)/(4"a"^2) - 2 - 6"a" + (3)/(2"a")`

= `(4"a"^2 + 1/(4"a"^2) - 2) - (6"a" - 3/(2"a"))`

= `(2"a" - 1/(2"a"))^2 - 3(2"a" - 1/(2"a"))`

= `(2"a" - 1/(2"a")) (2"a" - 1/(2"a") - 3)`

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अध्याय 5: Factorisation - Exercise 5.3

APPEARS IN

फ्रैंक Mathematics [English] Class 9 ICSE
अध्याय 5 Factorisation
Exercise 5.3 | Q 4.2

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