We are given that f(a,b)=2a−3b2+7−5a2+b3 and f(k,−3)=−10. We want to find k.
To find k, we substitute a=k and b=−3 into the expression for f(a,b):
f(k,−3)=2(k)−3(−3)2+7−5(k)2+(−3)3
Simplifying the right side:
f(k,−3)=2k−27+7−5k2+27
Combining like terms:
f(k,−3)=−5k2+2k
We are given that f(k,−3)=−10, so we can set this equal to −10 and solve for k:
−5k2+2k=−10
Move all the terms to one side:
−5k2+2k+10=0
Factor the expression:
$ -(5k^2 - 2k - 10) = 0$
(k+2)(5k−5)=0
Therefore, either k+2=0 or 5k−5=0. Solving for k in each case gives us k=−2 or k=1.
However, we are given that f(k,−3)=−10. Let's evaluate f at each possible value of k:
If k=−2, then f(−2,−3)=2(−2)−3(−3)2+7−5(−2)2+(−3)3=16.
If k=1, then f(1,−3)=2(1)−3(−3)2+7−5(1)2+(−3)3=−10.
Since only f(1,−3)=−10, the solution for k is k=1.
