For all real numbers x and y, the custom operation \blacktriangle is defined by x \blacktr… — GRE General Test

For all real numbers $x$ and $y$ , the custom operation $\blacktriangle$ is defined by $x \blacktriangle y = 2x^2 - y^2 + xy$ . The function $g$ is defined for all real numbers $n$ by $g(n) = n \blacktriangle 3$ . If $k$ is a positive constant such that $g(k) = 0$ , what is the value of $k$ ?

A
$\frac{2}{3}$
B
$1$
$\frac{3}{2}$ Answer
D
$3$
E
$9$

Answer

\frac{3}{2}

Evaluating

g(k) = k \blacktriangle 3

using the rule

x \blacktriangle y = 2x^2 - y^2 + xy

gives

2k^2 - 3^2 + 3k = 2k^2 + 3k - 9

. Setting this quadratic equal to zero gives

(2k - 3)(k + 3) = 0

. The roots are

k = \frac{3}{2}

and

k = -3

. Because

k

is specified to be positive, the only valid solution is

\frac{3}{2}

Step-by-Step Solution

Apply the definition of the custom operation

\blacktriangle

to evaluate

g(k)

g(k) = k \blacktriangle 3 = 2(k)^2 - (3)^2 + k(3) = 2k^2 - 9 + 3k = 2k^2 + 3k - 9

Substitute

x = k

and

y = 3

into the given expression

x \blacktriangle y = 2x^2 - y^2 + xy

Set the expression for

g(k)

equal to

0

and factor the quadratic equation.

2k^2 + 3k - 9 = (2k - 3)(k + 3) = 0

We need to find values of

k

for which

g(k) = 0

Solve for

k

and apply the constraint that

k

is a positive constant.

2k - 3 = 0 \implies k = \frac{3}{2}

k + 3 = 0 \implies k = -3

. Since

k > 0

k = \frac{3}{2}

Disregard the negative root

k = -3

because the problem specifies that

k

must be positive.

Key Concept

Combining custom symbol operations with function notation to solve quadratic equations

Estimated Time:1m 30s

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