मराठी

(i) 4⁢x^2 − 5⁢√2x − 7 = (2⁢√2x + 7)⁢(√2x − 1) (ii) x^2 + x + px + p = (x + 1)(x + p) - Mathematics

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प्रश्न

(i) `4x^2 - 5sqrt(2)x - 7 = (2sqrt(2)x + 7)(sqrt(2)x - 1)`

(ii) x2 + x + px + p = (x + 1)(x + p)

पर्याय

  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

MCQ
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उत्तर

Only (ii)

Explanation:

(i) The expression `4x^2 - 5sqrt(2)x - 7` can be factorized as:

`4x^2 - 5sqrt(2)x - 7 = (2sqrt(2)x + 7)(sqrt(2) x - 1)`

This matches the given factorization exactly.

(ii) The expression x2 + x + px + p factors as:

x2 + x + px + p = x2 + (1 + p)x + p

If factored as (x + 1)(x + p) = x2 + (p + 1)x + p, it matches the above only if (p) is a constant and the middle term is (p + 1)x.

Since the given expression has (x + px), which is x(1 + p), the factorization is correct only if (p) is a constant.

The expression x2 + x + px + p is equivalent to (x + 1)(x + p).

This confirms (ii) is also a valid factorization.

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पाठ 4: Factorisation - Exercise 4F [पृष्ठ ९१]

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नूतन Mathematics [English] Class 9 ICSE
पाठ 4 Factorisation
Exercise 4F | Q 4. | पृष्ठ ९१
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