Question

Consider the quadratic equation x^2=4x-5.How many solutions does the equation have?

Answer 1

Hello from MrBillDoesMath!

Answer:

2 (complex) solutions

x = 2 + i and x 2 - i

Discussion:

Rewrite x^2 = 4x - 5 as

x^2 -4x + 5 = 0

The solutions from the quadratic formula are

x = ( -(-4) +\- sqrt ( (-4)^2 - 4 (1)(5)) ) /2

x = ( 4 +\- sqrt(16-20)) )/2

x = ( 4 +\- sqrt(-4) ) /2

x = (4 +\- 2i) /2

x = 2 +\- i

Thank you,

MrB

Answer 2

It does not have a solution.

Explanation:

To solve this you just need to arrange the equation into the general formula:

x^2-4x+5=0

So we just insert that into the general formula:

`x=(-b+√(b^2-4ac) )/(2a)`

Now you just have to insert the values into this formula:

`x=(-b+√(b^2-4ac) )/(2a)\\x=(-(-4)+√(16-4(5)) )/(2*1)\\x=(+4+√(-20) )/(2)`

Since the square root is a negative number, you won´t be able to do that, so there is no solution for the equation.