α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α +

1.	α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α + 2-0 is 2, Find the value of theparameter αí thn rants of the quadratic equation a x2bxc 0, show that
Answer» assume the roots ofx^2 + ax + a + 2 = 0 to be t and 2t. So, here, Sum of roots = -(a)/1 => t + 2t = -a => 3t = -a => t = -a/3 Product of roots = (a + 2)/1 => t * 2t = (a + 2) => 2t^2 = (a + 2) Now, we'll substitute the value of t we found from by using sum of roots formula, => 2(-a/3)^2 = (a + 2) => 2a^2/9 = (a + 2) => 2a^2 = 9a + 18 => 2a^2 - 9a - 18 = 0 This is again a Quadratic equation, and can be easily solved by splitting the middle term. => 2a^2 - 12a + 3a - 18 = 0 => 2a(a - 6) + 3(a - 6) = 0 => (2a + 3)(a - 6) = 0 => (2a + 3) = 0; (a - 6) = 0 => a = -3/2; a = 6. Hence, the given condition can be satisfied by two values of a. a= -3/2, and a = 6. These values of a are the answer.

α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α + 2-0 is 2, Find the value of theparameter αí thn rants of the quadratic equation a x2bxc 0, show that

Answer»

assume the roots ofx^2 + ax + a + 2 = 0 to be t and 2t.

So, here,

Sum of roots = -(a)/1

=> t + 2t = -a

=> 3t = -a

=> t = -a/3

Product of roots = (a + 2)/1

=> t * 2t = (a + 2)

=> 2t^2 = (a + 2)

Now, we'll substitute the value of t we found from by using sum of roots formula,

=> 2(-a/3)^2 = (a + 2)

=> 2a^2/9 = (a + 2)

=> 2a^2 = 9a + 18

=> 2a^2 - 9a - 18 = 0

This is again a Quadratic equation, and can be easily solved by splitting the middle term.

=> 2a^2 - 12a + 3a - 18 = 0

=> 2a(a - 6) + 3(a - 6) = 0

=> (2a + 3)(a - 6) = 0

=> (2a + 3) = 0; (a - 6) = 0

=> a = -3/2; a = 6.

Hence, the given condition can be satisfied by two values of a. a= -3/2, and a = 6. These values of a are the answer.