Answer:
B
Step-by-step explanation:
y=x^(2)+4
y=4x
lets solve by substitution
4x=x^2+4
0=x^2-4x+4
0=(x-2)(x-2)
x=2 or x=2
y=4x
y=4*2
y=8
same for the other answer
(2,8)
this system has only one solution: (2,8)
Thus, the correct answer is B (one solution)
The system of equations y = x² + 4 and y = 4x has only one solution option (B) is correct.
Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.
As we know, the formula for the roots of the quadratic equation is given by:
[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]
We have:
y = x² + 4
y = 4x
Plug y = 4x in the first equation:
4x = x² + 4
x² - 4x + 4 = 0
After solving the above quadratic equation:
(x - 2)² = 0
x =2
y = 8
The solution is (x, y) = (2, 8)
Thus, the system of equations y = x² + 4 and y = 4x has only one solution option (B) is correct.
Learn more about quadratic equations here:
brainly.com/question/2263981
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