Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer: I. \(\sqrt {{x^2}\; + \;12} \; + \;\frac{7x}{{\sqrt {{x^2}\; + \;12} }} = 2\sqrt {{x^2}\; + \;12} \) II. 2y2 – 17y + 36 = 0
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
I. \(\sqrt {{x^2}\; + \;12} \; + \;\frac{7x}{{\sqrt {{x^2}\; + \;12} }} = 2\sqrt {{x^2}\; + \;12} \)
II. 2y2 – 17y + 36 = 0
1). if x > y
2). if x ≥ y
3). if x < y
4). if x ≤ y
1 answers
I. $(\sqrt {{x^2}\; + \;12} \; + \;\frac{7x}{{\sqrt {{x^2}\; + \;12} }} = 2\sqrt {{x^2}\; + \;12} )$
$(\Rightarrow \;\frac{{{x^2}\; + \;12\; + \;7x}}{{\sqrt {{x^2}\; + \;12} }} = 2\sqrt {{x^2}\; + \;12} )$
⇒ x2 + 12 + 7x = 2(x2 + 12)
⇒ x2 – 7x + 12 = 0
⇒ x2 – 4x - 3x + 12 = 0
⇒ x(x – 4) - 3(x – 4) = 0
⇒ (x – 4)(x - 3) = 0
∴ x = 4 or x = 3
II. 2y2 – 17y + 36 = 0
⇒ 2y2 – 9y – 8y + 36 = 0
⇒ y(2y – 9) – 4(2y – 9) = 0
⇒ (2y – 9)(y – 4) = 0
∴ y = 9/2 or y = 4
When x = 4, x < y for y = 9/2 and x = y for y = 4
And when x = 3, x < y for y = 9/2 and x = y for y = 4
∴ x ≤ yOther Questions
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