Q10. Given that the roots of the equation 2x^{2}=-2x+3 are a and 3. The equations whose roots are 1/a. and 1/b is

Answer 1

To find the equation whose roots are 1/a and 1/b, we first need to find the values of a and b.

We are given that the equation 2x^2 + 2x - 3 = 0 has roots a and 3. To find these roots, we can set the equation equal to 0 and use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In our case, a = 2, b = 2, and c = -3. Plugging these values into the quadratic formula, we get:

a = (-2 ± √(2^2 - 4(2)(-3)))/(2(2))

Simplifying under the square root:

a = (-2 ± √(4 + 24))/(4)

a = (-2 ± √(28))/(4)

a = (-2 ± 2√(7))/(4)

Simplifying the expression inside the square root:

a = (-1 ± √(7))/(2)

So we have two possible values for a: (-1 + √(7))/2 and (-1 - √(7))/2.

Next, we can find the reciprocal of each value of a to find the roots 1/a and 1/b:

1/a = 1/((-1 + √(7))/2) = 2/(-1 + √(7))
1/b = 1/((-1 - √(7))/2) = 2/(-1 - √(7))

Therefore, the equation with roots 1/a and 1/b is:

2x^2 + (1 + √(7))x + (1 - √(7)) = 0

or

2x^2 + (1 - √(7))x + (1 + √(7)) = 0