Question

Out of the 2 roots of x^2+(1-2y)x+(y^2-y-2)=0 one root is greater than 3 and the other root is less than 3, then the limits of y are​

Answer 1

Answer:

Hello!

The answer is y belongs to (2,5)

Step-by-step explanation:

This is the given equation-

x^2+(1-2y)x+(y^2-y-2)=0

Its roots are-

[-b+(b^2-4ac)^1/2]/2a and [-b-(b^2-4ac)^1/2]/2a

here,

a = 1, b=(1+2y) and c=(y^2-y-2)

You will get its roots as-

y-2 and y+1

As the former is lesser than the latter,

y-2<3 and y+1>3

Therefore,

y<5 and y>2

y belongs to the set (2,5)

Hope this helps!

Answer 2

Answer:

2<y<5

Step-by-step explanation:

The graph of given quadratic opens upwards. For one root to be greater than 3 and other less than 3, f(3)<0.

This gives,

⇒3^2+(1-2y)3+y^2-y-2<0

⇒y^2-7y+10<0

⇒(y-2)(y-5)<0

⇒  y<5    y>2

⇒2<y<5