# the quadratic equation x^2 +2(1+λ)x +(5+2λ)=0 has roots alpha and beta find the set of values of λfor which the roots are real .

the quadratic equation x^2 +2(1+λ)x +(5+2λ)=0 has roots alpha and beta

find the set of values of λfor which the roots are real .

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x^2+2(1+v)x +(5+2v) =0

To find the roots:

x1= 1+ sqrt(1-4(5+2v))

= 1+ sqrt(1-20-8v) = 1+sqrt(-19-8v)

The function has solution iff sqrt(-19-8v) is defined . which means that sqrt(-19-8v) does not have a negative value:

==> sqrt(-19 -8v) =>0

==> -19-8v => 0

==> -8v => 19

==> v => -19/8

==> v belongs to the interval [-19/8,inf)

x^2+2(1+l)x+(5+2l) = 0

To find the values for which the roots are reak:

Solution:

We know tht the roots are real for the quadratic equation ax^2+bx+c = 0 , if the discriminant , b^2- 4ac > 0.

So inthe present case , a= 1, b = 2(1+l) and c = 5+2l. So the the discriminant {2(1+l)}^2 - 4*1*(5+2l) > 0. Or

4(1+l)^2-4(5+2l) > 0. Divide by 4.

(1+l)^2 - (5+2l) > 0.

1+2l+l^2-5-l > 0

l^2 -4 > 0. Or

(l+2)(l-2) > 0.

So both factors need to be positive or bot factors should be - ve in order that the product is positive.

So

l > 2 Or l <-2 makes the equation to have real roots.