Respuesta :
Hello /
f(x) = x²+7 (color red) : f(x) =0 no solution
g(x) = 2x² (color bleue) : g(x) = 0 one solution
h(x) = (1/4)x² -4 ( color green ) : h(x) = 0 two solutions
f(x) = x²+7 (color red) : f(x) =0 no solution
g(x) = 2x² (color bleue) : g(x) = 0 one solution
h(x) = (1/4)x² -4 ( color green ) : h(x) = 0 two solutions
Answer:
We have to find the number of solutions of each of the following equations by observing their graphs.
We know that the number of real solutions are the number of times the graph touches the x-axis. i.e. the point at which the function is zero.
1)
[tex]x^2+7=0[/tex]
Clearly this graph has no real solutions.
Also we can see from the equation that the function takes no real value such that the function is zero.
since,
[tex]x^2\geq 0\\\\x^2+7>0[/tex]
Hence, it has no real solution.
2)
[tex]3x^2=0[/tex]
This graph has one real solution i.e. 0 with multiplicity 2.
Also it could be seen as:
[tex]3x^2=0\\\\x=0,0[/tex]
are the possible solution of the equation.
Hence, the function has one real solution with multiplicity 2.
3)
[tex]\dfrac{1}{4}x^2-4=0[/tex]
From the graph of this function we see that it has 2 real solutions i.e. 4 and -4.
( Also it could be seen as:
[tex]\dfrac{1}{4}x^2=4\\\\x^2=16\\\\x=\pm4[/tex] )