Question

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

REALLY NEED HELP WITH THIS ONE

Answer 1

Explanation:

Answer 2

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)
5x²+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