मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

Solve y+1+2y-5 = 3 - Mathematics

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प्रश्न

Solve `sqrt(y + 1) + sqrt(2y - 5)` = 3

बेरीज

उत्तर

`sqrt(y + 1) + sqrt(2y - 5)` = 3           ...(squaring on both sides)

`(sqrt(y + 1) + sqrt(2y - 5))^2` = 32 

`(sqrt(y + 1))^2 + (sqrt(2y - 5))^2 + 2(sqrt(y + 1))(sqrt(2y - 5))` = 9

`y + 1 + 2y - 5 + 2 sqrt((y + 1)(2y - 5))` = 9

`3y - 4 + 2 sqrt(2y^2 - 5y + 2y - 5)` = 9

`2sqrt(2y^2 - 3y - 5)` = 9 + 4 – 3y

= 13 – 3y                  ...(squaring on both sides)

`[2sqrt(2y^2 - 3y - 5)]^2` = (13 – 3y)2

4(2y2 – 3y – 5) = 169 + 9y2 – 78y

8y2 – 12y – 20 = 169 + 9y2 – 78y

8y2 – 9y2 – 12y + 78y – 20 – 169 = 0

–y2 – 66y – 189 = 0

y2 – 66y + 189 = 0

(y – 3) (y – 63) = 0

y = 3 or y = 63

The value of y is 3 and 63.

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पाठ 3: Algebra - Unit Exercise – 3 [पृष्ठ १५७]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 3 Algebra
Unit Exercise – 3 | Q 10 | पृष्ठ १५७
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