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 =…
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
Substitute (x = 4) into (x^{2}+bx + 16=0). We get (4^{2}+4b+16 = 0).
Step2: Simplify the equation
Calculate (4^{2}=16), so 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