QUESTION: Find the quadratic equation,which has difference of roots equal to 2 and the difference of the squares of roots equal to 5
(x-r1)(x-r2)=0
now given r1-r2=2
and r1^2-r2^2=5
from the last equation
(r1-r2)(r1+r2)=5
2(r1+r2)=5
r1+r2=2.5 but
r1-r2=2
adding equation..
2r1=4.5, so now you have r1, and you can compute r2.
r1=2.25
r2=0.25
check squares..
2.25^2-.25^2= yep, 5
To find the quadratic equation, let's assume the roots of the equation are denoted as "x" and "y".
According to the given information, we know that the difference between the roots is 2. Therefore, one root must be x, and the other must be x + 2.
The difference between the squares of the roots is given as 5. Mathematically, this can be represented as (x + 2)^2 - x^2 = 5.
Expanding the equation, we get x^2 + 4x + 4 - x^2 = 5.
The x^2 terms cancel each other out, leaving us with 4x + 4 = 5.
Subtracting 4 from both sides, we have 4x = 1.
Finally, dividing both sides by 4, we find that x = 1/4.
Now that we know one of the roots, we can find the other root by adding 2 to x. Therefore, the other root would be 1/4 + 2 = 9/4.
Now we have both roots, x = 1/4 and y = 9/4.
To construct the quadratic equation, we know that the sum of the roots is equal to -b/a and the product of the roots is equal to c/a.
The equation can be written as x^2 - (sum of roots)x + product of roots = 0.
Plugging in the values, the quadratic equation becomes:
(x - 1/4)(x - 9/4) = 0.
Expanding this equation, we get x^2 - (10/4)x + (9/16) = 0.
Finally, multiplying the whole equation by 16, we get:
16x^2 - 40x + 9 = 0.
So, the quadratic equation with roots that have a difference of 2 and a difference of the squares of roots equal to 5 is 16x^2 - 40x + 9 = 0.