Why are and Iby 4and - 1 zeroes of the polynomial p(x) = 4² +3x-1?
1 answer:
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Answer: the system has no solution.
Step-by-step explanation:
\displaystyle\\
\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\
Multiply\ both\ sides\ of\ the\ equation\ (2)\ by\ y\ (y\neq 0):\\
x^2y+4y^2+16y=0\\
We\ substitute\ equation\ (1)\ into\ equation\ (2):\\
16+4y^2+16y=0\\
4y^2+16y+16=0\\
4*(y^2+4y+4)=0\\
4*(y^2+2*y*2+2^2)=0\\
4*(y+2)^2=0\\
Divide\ both\ sides\ of\ the \ equation\ by\ 4:\\
(y+2)^2=0\\
(y+2)*(y+2)=0\\
So,\ y+2=0\\
y=-2.\\