If a and b are the two real solutions to the quadratic equation 2x^2 - 11x + 12 = 0, such … — GRE General Test

If $a$ and $b$ are the two real solutions to the quadratic equation $2x^2 - 11x + 12 = 0$ , such that $a > b$ , what is the value of $a - 2b$ ?

Answer: 1

The value of

a - 2b

is 1.

Factoring

2x^2 - 11x + 12 = 0

yields

(2x - 3)(x - 4) = 0

, giving solutions

x = 1.5

and

x = 4

. Given that

a > b

, we must set

a = 4

and

b = 1.5

. Substituting these values into

a - 2b

gives

4 - 2(1.5) = 1

Factor the quadratic equation

2x^2 - 11x + 12 = 0

(2x - 3)(x - 4) = 0

Find two linear factors whose product expands to

2x^2 - 11x + 12

Find the roots of the equation

x = \frac{3}{2} = 1.5

and

x = 4

Apply the zero product property:

2x - 3 = 0 \Rightarrow x = 1.5

and

x - 4 = 0 \Rightarrow x = 4

Assign values to

a

and

b

based on the inequality

a > b

a = 4

and

b = 1.5

Since

4 > 1.5

a

must be 4 and

b

must be 1.5.

Evaluate the targeted expression

a - 2b

4 - 2(1.5) = 1

Substitute

a = 4

and

b = 1.5

into

a - 2b

Factoring Quadratic Equations

Question