If one root of the equation 2x^2-3x+b is double the other, what is b?

Let’s call the roots of the quadratic equation 2x^2 – 3x + b as r₁ and r₂. According to the given information, one root is double the other. This means that we can express r₂ as 2r₁.

Now, let’s use Vieta’s formulas to relate the coefficients of the quadratic equation to its roots. Vieta’s formulas state that for a quadratic equation of the form ax^2 + bx + c = 0 with roots r₁ and r₂, the following relationships hold:

- The sum of the roots is equal to the negation of the coefficient of the linear term, b: r₁ + r₂ = -b/2a
- The product of the roots is equal to the constant term, c, divided by the coefficient of the quadratic term, a: r₁ * r₂ = c/a

In our equation, a = 2, and we already know that r₂ = 2r₁. So, let’s use these relationships:

From the first relationship: r₁ + 2r₁ = -(-3) / (2 * 2)

3r₁ = 3/4

Now, divide both sides by 3 to solve for r₁: r₁ = (3/4) / 3 r₁ = 1/4

Now that we have found the value of r₁, we can find r₂, as r₂ is double r₁: r₂ = 2 * (1/4) r₂ = 1/2

Finally, we can use the second relationship to find b: r₁ * r₂ = b / 2

(1/4) * (1/2) = b / 2

(1/8) = b / 2

Now, multiply both sides by 2 to solve for b: b = 2 * (1/8)

b = 1/4

So, the value of b that satisfies the condition where one root of the equation 2x^2 – 3x + b is double the other is b = 1/4.