Cummings Bright
04/06/2023 · Senior High School
If \( x=1 \) and \( y=6 \) is a solution of the equation \( 8 x-k y+k^{2}=0 \), find the value of \( k \).
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Step-by-step Solution
To find the value of \( k \) given that \( x = 1 \) and \( y = 6 \) is a solution of the equation
\[
8x - ky + k^2 = 0,
\]
we can substitute \( x = 1 \) and \( y = 6 \) into the equation.
Substituting these values, we have:
\[
8(1) - k(6) + k^2 = 0.
\]
This simplifies to:
\[
8 - 6k + k^2 = 0.
\]
Rearranging the equation gives us:
\[
k^2 - 6k + 8 = 0.
\]
Next, we can solve this quadratic equation using the quadratic formula:
\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 1 \), \( b = -6 \), and \( c = 8 \). Plugging in these values, we get:
\[
k = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}.
\]
Calculating the discriminant:
\[
(-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4.
\]
Now substituting back into the formula:
\[
k = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}.
\]
This gives us two possible solutions for \( k \):
\[
k = \frac{6 + 2}{2} = \frac{8}{2} = 4,
\]
and
\[
k = \frac{6 - 2}{2} = \frac{4}{2} = 2.
\]
Thus, the possible values of \( k \) are \( 4 \) and \( 2 \).
To summarize, the values of \( k \) that satisfy the equation are:
\[
\boxed{2} \quad \text{and} \quad \boxed{4}.
\]
Quick Answer
The values of \( k \) are 2 and 4.
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