Math, asked by ashutosh1123, 1 year ago

16( ii) please solve this.....

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Answers

Answered by nalinsingh
0

Hey!

Refer the below attachment !

Thanks!

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ashutosh1123: Its wrong
nalinsingh: It is absolutely correct.
nalinsingh: I have checked 2 times.
ashutosh1123: No its wrong
Answered by siddhartharao77
3

Answer:

6,5

Step-by-step explanation:

(ii)

Given: √3x² - 7x - 30 - √2x² - 7x - 5 = x - 5

⇒ √3x² - 7x - 30 = (x - 5) + √2x² - 7x - 5

On Squaring both sides, we get

⇒ (√3x² - 7x - 30)² = [(x - 5) + (√2x² - 7x - 5)]²

∴ (a + b)² = a² + b² + 2ab

⇒ 3x² - 7x - 30 = [(x² + 25 - 10x) + (2x² - 7x - 5) + 2(x - 5)(√2x² - 7x - 5)

⇒ 3x² - 7x - 30 = [x² + 25 - 10x + 2x² - 7x - 5 + 2x(√2x² - 7x - 5) - 10(√2x² - 7x - 5)]

⇒ 3x² - 7x - 30 = [3x² + 2x(√2x² - 7x - 5) - 17x - 10(√2x² - 7x - 5) + 20]

⇒ 3x² - 7x - 30 - 3x² - 2x(√2x² - 7x - 5) + 17x + 10(√2x² - 7x - 5) - 20 = 0

⇒ 10x - 2x(√2x² - 7x - 5) + 10(√2x² - 7x - 5) - 50 = 0

⇒ 2[5x - x(√2x² - 7x - 5) + 5(√2x² - 7x - 5) - 25] = 0

⇒ [5x - (√2x² - 7x - 5) + 5(√2x² - 7x - 5) - 25] = 0

⇒ [(-x√2x² - 7x - 5 + 5x) + (5√2x² - 7x - 5) - 25] = 0

⇒ 5[(√2x² - 7x - 5) - 5)] - [x(√2x² - 7x - 5) - 5] = 0

⇒ -[(√2x² - 7x - 5) - 5](x - 5) = 0

(i)

(√2x² - 7x - 5) - 5 = 0

⇒ √2x² - 7x - 5 = 5

On Squaring both sides, we get

⇒ 2x² - 7x - 5 = 25

⇒ 2x² - 7x - 30 = 0

⇒ 2x² + 5x - 12x - 30 = 0

⇒ x(2x +5) - 6(2x + 5) = 0

⇒ (x - 6)(2x + 5) = 0

⇒ x = 6, x = -5/2{Neglect}

⇒ x = 6.

(ii)

x - 5 = 0

x = 5.

∴ The values of x are 6 and 5.

Hope it helps!


ashutosh1123: Thanks
siddhartharao77: Welcome
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