For each equation, determine the number of solutions and place on the appropriate field in the table.

REALLY NEED HELP WITH THIS ONE

For each equation determine the number of solutions and place on the appropriate field in the table REALLY NEED HELP WITH THIS ONE class=

Respuesta :

we know that
In a quadratic equation 
ax²+bx+c=0
the term (b² - 4ac) is called the "discriminant" because it can "discriminate" between the possible types of answer:
so
(b² - 4ac) > 0 we get two real solutions
(b² - 4ac) =0  we get just ONE real solution
(b² - 4ac) < 0 we get complex solutions or no real solutions

part 1)
5x
²+2=4x
5x²-4x+2=0
a=5
b=-4
c=2
(b² - 4ac)=(-4)²-4*(5)*(2)------> 16-40-----> -24
so
(b² - 4ac) < 0-------> No Real Solutions

the answer Part 1) is
No Real Solutions

part 2) 
3(x+5)²=-2
3(x²+10x+25)=-2
3x²+30x+75=-2
3x²+30x+77=0
a=3
b=30
c=77
(b² - 4ac)=(30)²-4*(3)*(77)------> 900-924 -----> -24
so
(b² - 4ac) < 0-------> No Real Solutions

the answer Part 2) is
No Real Solutions

part 3) 
4x²-16x=0
a=4
b=-16
c=0
(b² - 4ac)=(-16)²-4*(4)*(0)------> 256 
so
(b² - 4ac) > 0-------> Two Real Solutions

the answer Part 3) is
Two Real Solutions

part 4) 
3x²+24x=-48
3x²+24x+48=0
a=3
b=24
c=48
(b² - 4ac)=(24)²-4*(3)*(48)------> 0
so
(b² - 4ac) =0 -------> One Real Solution

the answer Part 4) is
One Real Solution

Answer:

Step-by-step explanation:

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