Keerthana
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I. 2x27x60=02{x^2} - 7x - 60 = 0

II. 3y2+13y+4=03{y^2} + 13y + 4 = 0

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

Answer : Option E
Explanation :

I. 2x27x60=02{{ m{x}}^2} - 7{ m{x}} - 60 = 0

2x215x+8x60=02{{ m{x}}^2} - 15{ m{x}} + 8{ m{x}} - 60 = 0

x(2x15)+4(2x15)=0{ m{x}}left( {2{ m{x}} - 15} ight) + 4left( {2{ m{x}} - 15} ight) = 0

(x+4)(2x15)=0left( {{ m{x}} + 4} ight)left( {2x - 15} ight) = 0

x=4,152x = - 4,frac{{15}}{2}II. 3y2+13y+4=03{{ m{y}}^2} + 13{ m{y}} + 4 = 0

3y2+12y+y+4=03{{ m{y}}^2} + 12{ m{y}} + { m{y}} + 4 = 0

3y(y+4)+1(y+4)=03{ m{y}}left( {{ m{y}} + 4} ight) + 1left( {{ m{y}} + 4} ight) = 0

(3y+1)(y+4)=0left( {3{ m{y}} + 1} ight)left( {{ m{y}} + 4} ight) = 0

y=13,4y = - frac{1}{3}, - 4

no relation can be established

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