हिंदी

Solve the following equation by factorization: sqrt(2x + 9) = (13 - x)

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प्रश्न

Solve the following equation by factorization:

`sqrt(2x + 9) = (13 - x)`

Solve for x: 

`sqrt(2x + 9) + x = 13`

योग
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उत्तर

Given,

⇒ `sqrt(2x + 9) = (13 - x)`

Or Given, 

⇒ `sqrt(2x + 9) + x = 13`

⇒ `sqrt(2x + 9) = 13 - x`

Squaring both sides we get:

⇒ (2x + 9) = (13 – x)2 

⇒ 2x + 9 = (132) + (x)2 – 2 × 13 × x

⇒ 2x + 9 = 169 + x2 – 26x

⇒ x2 – 26x + 169 – 2x – 9 = 0

⇒ x2 – 28x + 160 = 0

⇒ x2 – 20x – 8x + 160 = 0

⇒ x(x – 20) – 8(x – 20) = 0

⇒ (x – 20)(x – 8) = 0

⇒ (x – 20) = 0 or (x – 8) = 0   ...[Using zero-product rule]

⇒ x = 20 or x = 8

⇒ x = 8

Substituting x = 20 in the L.H.S. of this equation `sqrt(2x + 9) = (13 - x)`

⇒ `sqrt(2(20) + 9)`

⇒ `sqrt(40 + 9)`

⇒ `sqrt(49)`

⇒ 7

Substituting x = 20 in the R.H.S. of this equation `sqrt(2x + 9) = (13 - x)`

⇒ 13 – x

⇒ 13 – 20

⇒ –7

L.H.S ≠ R.H.S.

∴ x = 20 is not valid.

Hence, x = {8}.

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Notes

Students should refer to the solution part of "Given" according to their solution. Rest of the solution is same.

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Quadratic Equation - EXERCISE 5A [पृष्ठ ५७]

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 5 Quadratic Equation
EXERCISE 5A | Q 43. | पृष्ठ ५७
सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 5 Quadratic Equations
EXERCISE 5 (B) | Q 15. (iv) | पृष्ठ ५४
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