If x is a positive real number such that 2x^2 - 5x - 3 = 0, what is the value of x^2 + 1? — GRE General Test

If $x$ is a positive real number such that $2x^2 - 5x - 3 = 0$ , what is the value of $x^2 + 1$ ?

Factoring the quadratic equation

2x^2 - 5x - 3 = 0

yields

(2x + 1)(x - 3) = 0

, giving solutions

x = -\frac{1}{2}

and

x = 3

. Because

x

is specified as a positive real number, we choose

x = 3

. Substituting

x = 3

into

x^2 + 1

yields

3^2 + 1 = 10

Factor the quadratic equation

2x^2 - 5x - 3 = 0

(2x + 1)(x - 3) = 0

Find two linear factors whose product gives

2x^2 - 5x - 3

Solve for the roots of the equation.

x = -\frac{1}{2}

x = 3

Set each factor equal to zero:

2x + 1 = 0 \implies x = -\frac{1}{2}

and

x - 3 = 0 \implies x = 3

Apply the constraint

x > 0

to identify the valid root.

x = 3

The root

x = -\frac{1}{2}

is negative and thus violates the given condition that

x

is positive.

Substitute

x = 3

into the target expression

x^2 + 1

3^2 + 1 = 9 + 1 = 10

Evaluate the expression using the valid positive value of

x

Solving quadratic equations by factoring and applying domain constraints

Estimated Time:45s

Question