Non-Real Numbers in Quadratic Formulas?

Question

Would (x-3)^2 = 5 turn out to be a non-real number answer by solving using the quadratic formula?

I know you can solve it by square roots .. how about factoring?

Answer 1

No, it is a real number.

To factor this write it as

(x - 3)^2 - 5

This is the difference of 2 squares, (x - 3) and sqrt(5). So the factorization is

((x - 3) + sqrt(5))*((x - 3) - sqrt(5))

In this form it is easy to see if the answer is real or not, if the right side is nonnegative there are real solutions, if it is < 0 then there are none.

Answer 2

To check if a quadratic equation, ax^2 +bx + c = 0, has real roots, you need to calculate the value of its discriminant. The discriminant is the name given to the expression b^2 - 4ac.
If the discriminant is less than zero then there are no real solutions. If it is equal to zero there is one real ('repeated') solution and if greater than zero then two real solutions.

In this case, if you multiply out the brackets and rearrange the equation into the above form, you get x^2-6x+4=0 and so
a=1, b=-6 and c=4
Therefore b^2-4ac = 36-4=32
Since the discriminant is greater than zero this equation has two real solutions. Using the quadratic formula will give you these two solutions. The solutions to this particular equation are irrational numbers and therefore it cannot be solved by factoring. Any quadratic equation can be solved using the formula, so if in doubt use it! In this case the solutions are:
x=3+sqrt5 and x=3-sqrt5

Answer 3

the solution wouldn't be imaginary. it doesn't factor out perfectly, but when you solve it using the quadratic formula you don't get a negative square root.

Answer 4

(x-3)^2 = 5

Subtract both sides by (√5)^2
(x-3)^2-( √5)^2=0

useing　a^2-b^2=(a+b)(a-b)
{(x-3)+( √5)}{(x-3)-( √5)}=0
{x-(3-√5)}{x-(3+√5)}=0

x=3+√5, 3-√5

they are Real Number.


If,
(x-3)^2 = -5

In the same way but useing i,

Subtract both sides by (i√5)^2
(x-3)^2-( i√5)^2=0

useing　a^2-b^2=(a+b)(a-b)
{(x-3)+( i√5)}{(x-3)-( i√5)}=0
{x-(3-i√5)}{x-(3+i√5)}=0

x=3+i√5, 3-i√5

Answer 5

(x - 3)^2 = 5
x - 3 = sqrt(5)
x = 3 ± sqrt(5)

for better help, go to www.quickmath.com

Answer 6

x^2-6x+9=5
x^2-6x+4=0
can't be factored
6 +- sqr (36-4(1)(4))/2
ans:5.236 and .763

Answer 7

Damn you are smart to have that lol but im not im sorry