SOLUTION: f(x)=x^2+6x+20+k(x^2-3x-12) where k is real. Find value of k given that f(x) is a straight line
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Question 1205621: f(x)=x^2+6x+20+k(x^2-3x-12) where k is real. Find value of k given that f(x) is a straight line
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
f(x)=x^2+6x+20+k(x^2-3x-12)
f(x)=x^2+6x+20+kx^2-3kx-12k
f(x)=(x^2+kx^2)+(6x-3kx)+(20-12k)
f(x)=(1+k)x^2+(6-3k)x+(20-12k)
f(x) is linear if and only if the x^2 coefficient is 0.
1+k = 0
k = -1 is the final answer.
If k = -1, then,
f(x)=x^2+6x+20+k(x^2-3x-12)
f(x)=x^2+6x+20-1(x^2-3x-12)
f(x)=x^2+6x+20-x^2+3x+12
f(x)=(x^2-x^2)+(6x+3x)+(20+12)
f(x)=0x^2+9x+32
f(x)=9x+32
which proves f(x) is linear when k = -1.
Answer: k = -1
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
f(x) is the equation of a straight line only if the coefficient of the x^2 term is zero. We don't care about the other terms.
The function is
f(x) = x^2 + ... + k(x^2+...) = (x^2 + kx^2) + ... = (1+k)x^2 + ...
So
1+k = 0
k = -1
ANSWER: -1
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