the only solution of the equation $x^{2}+bx + 16 = 0$ is $x = 4$. what is the value of $b$?\n$b=-16$\n$b=-8$\…
the only solution of the equation $x^{2}+bx + 16 = 0$ is $x = 4$. what is the value of $b$?\n$b=-16$\n$b=-8$\n$b = 8$\n$b = 16$
Answer
Explanation:
Step1: Substitute x = 4 into equation
Since x = 4 is a solution of $x^{2}+bx + 16=0$, we substitute x = 4 into the equation: $4^{2}+4b+16 = 0$.
Step2: Simplify the equation
Calculate $4^{2}=16$, then the equation becomes $16 + 4b+16=0$, which simplifies to $4b+32 = 0$.
Step3: Solve for b
Subtract 32 from both sides: $4b=-32$. Then divide both sides by 4: $b=\frac{-32}{4}=-8$.
Answer:
$b = - 8$