Q10. Given that the roots of the equation 2x^{2}=-2x+3 are Alpha and Beta. The equations whose roots are 1/alpha and 1/Beta is
Q11. if alpha and ẞ are the roots of the equation 3x^{2}-5x=-2, then \alpha^{2}+\beta^{2} is
Q10. If we have the equation 2x^2 + 2x - 3 = 0, then we can find the roots using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 2, and c = -3. Plugging in these values, we get:
x = (-2 ± √(2^2 - 4 * 2 * -3)) / (2 * 2)
= (-2 ± √(4 + 24)) / 4
= (-2 ± √28) / 4
= (-2 ± 2√7) / 4
= -1/2 ± √7/2
So the roots of the equation are (-1 + √7)/2 and (-1 - √7)/2.
Now, we want the roots of the equation 1/alpha and 1/beta. If r is a root of the equation Ax^2 + Bx + C = 0, then 1/r is a root of the equation Cx^2 + Bx + A = 0. So in this case, the roots of the equation 1/alpha and 1/beta are:
1/alpha = 1 / (-1 + √7)/2 = 2 / (-1 + √7)
1/beta = 1 / (-1 - √7)/2 = 2 / (-1 - √7)
Therefore, the equations whose roots are 1/alpha and 1/beta are:
2x - (-1 + √7) = 0
and
2x - (-1 - √7) = 0
Q11. If we have the equation 3x^2 - 5x + 2 = 0, then we can find the roots using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -5, and c = 2. Plugging in these values, we get:
x = (-(-5) ± √((-5)^2 - 4 * 3 * 2)) / (2 * 3)
= (5 ± √(25 - 24)) / 6
= (5 ± √1) / 6
= (5 ± 1) / 6
So the roots of the equation are 1 and 2/3.
Now, we want to find α^2 + ß^2. We know that (α + ß)^2 = α^2 + ß^2 + 2αß. From the quadratic equation, we know that (α + ß) = -(-5)/3 = 5/3, and αß = 2/3.
Plugging these values into the equation (α + ß)^2 = α^2 + ß^2 + 2αß, we get:
(5/3)^2 = α^2 + ß^2 + 2(2/3)
25/9 = α^2 + ß^2 + 4/3
25/9 - 4/3 = α^2 + ß^2
(25 - 12)/9 = α^2 + ß^2
13/9 = α^2 + ß^2
Therefore, α^2 + ß^2 is equal to 13/9.