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

For all positive real numbers $x$ and $y$ , the custom operation $\star$ is defined by $x \star y = \frac{x^2 + y^2}{xy}$ . The function $g$ is defined by $g(t) = 4(t \star 3) - 7$ . If $g(k) = 3$ , where $k > 3$ , what is the value of $k$ ?

Cevap: 6

Cevap

The value of

k

is 6.

Solving

g(k) = 3

yields

4(k \star 3) - 7 = 3

, so

k \star 3 = \frac{5}{2}

. Replacing

k \star 3

with

\frac{k^2+9}{3k}

results in the quadratic equation

2k^2 - 15k + 18 = 0

. The roots are

1.5

and

6

. Because

k

must be strictly greater than

3

, the only valid solution is

6

Adım Adım Çözüm

Isolate the custom operation term in the function equation

k \star 3 = \frac{5}{2}

Setting

g(k) = 4(k \star 3) - 7 = 3

allows us to find the required value of

k \star 3

by adding 7 to both sides and dividing by 4.

Express the custom operation algebraically using its definition

\frac{k^2 + 9}{3k} = \frac{5}{2}

By definition,

x \star y = \frac{x^2 + y^2}{xy}

, so

k \star 3 = \frac{k^2 + 3^2}{3(k)} = \frac{k^2 + 9}{3k}

Form and solve the quadratic equation

k = 1.5 \text{ or } k = 6

Cross-multiplying gives

2(k^2 + 9) = 5(3k)

, which simplifies to

2k^2 - 15k + 18 = 0

. Factoring as

(2k - 3)(k - 6) = 0

yields two roots:

k = 1.5

and

k = 6

Apply the domain constraint

k = 6

The problem specifies that

k > 3

. Since

1.5 \le 3

k = 1.5

is rejected, leaving

k = 6

as the unique correct answer.

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