If r_1 and r_2 are the two distinct real solutions to the quadratic equation x^2 - 4x - 21… — GRE General Test

If $r_1$ and $r_2$ are the two distinct real solutions to the quadratic equation $x^2 - 4x - 21 = 0$ , where $r_1 > r_2$ , what is the value of $r_1 + 2r_2$ ?

$1$ Answer
B
$13$
C
$-11$
D
$-5$
E
$17$

Answer

The correct value of

r_1 + 2r_2

1

Factoring the quadratic polynomial

x^2 - 4x - 21

gives

(x - 7)(x + 3) = 0

, which yields solutions

x = 7

and

x = -3

. Applying the constraint

r_1 > r_2

identifies

r_1 = 7

and

r_2 = -3

. Substituting these values into

r_1 + 2r_2

yields

7 + 2(-3) = 7 - 6 = 1

Step-by-Step Solution

Factor the quadratic equation

(x - 7)(x + 3) = 0

Find two numbers that multiply to

-21

and add up to

-4

, which are

-7

and

3

Solve for the roots of the equation

x = 7

x = -3

Set each linear factor equal to zero:

x - 7 = 0 \implies x = 7

and

x + 3 = 0 \implies x = -3

Assign values to

r_1

and

r_2

based on the condition

r_1 > r_2

r_1 = 7

and

r_2 = -3

Since

7 > -3

r_1

must be

7

and

r_2

must be

-3

Evaluate the expression

r_1 + 2r_2

7 + 2(-3) = 7 - 6 = 1

Substitute

r_1 = 7

and

r_2 = -3

into the given expression.

Key Concept

Solving quadratic equations by factoring and evaluating expressions involving roots.

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

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