Question

x^2-4x+14=49

Answer 1

ok so i think i have the concept on what to do next can you tell me if its right or wrong

Answer 2

\[x^2-4x-49=0\]

Answer 3

Subtract both sides by 49

Answer 4

it started out as.....

Answer 5

\[x^2-4x+4=49\]

Answer 6

i then subtracted 49 from both sides to have it set = to 0 is that right

Answer 7

No it is

x^2-4x+14=49

Now you subtract both sides by 49

Answer 8

im sorry that was a typo the equation started as x^2-4x+4=49

Answer 9

so whats next combining the like terms??

Answer 10

Subtracting both sides by 49

What do you get ?

Answer 11

x^2-4x+4-49=49-49

Answer 12

you would get x^2-4x-45=0

Answer 13

x^2-4x+14=49
x^2-4x=49-14
x^2-4x=35
x(x-4_=35
x=35 or x-4=35
x=39

Answer 14

line 4 edit: x(x-4)=35

Answer 15

@aleksas1234 That's not correct. Insert x = 39 into the equation, you'll see that it won't hold. Furthermore, you'll always get 2 solutions from a quadratic equation. Thats's proven, it's called the Fundamental Theorem of Algebra.

Answer 16

So, for the solution, lets start off at
\[x^2 - 4x - 45 = 0\]There are 2 ways to solve this. Have you heard of the solution formula for quadratic equations or linear factorization of polynomials?

Answer 17

@Stiwan owh i found my mistace, ty

Answer 18

For any quadratic equation of the form\[x²+px + q = 0 \] holds that
\[x_1 = - \frac{p}{2} + \sqrt{(\frac{p}{2})^2 - q}\]\[x_2 = -\frac{p}{2} - \sqrt{(\frac{p}{2})^2 - q}\]
Where x_1, x_2 are the zeroes of the equation above. By plugging in p = -4 and q = -5
You get x_1 = 9, x_2 = -5