What is (x-3)^2 = 5 in ax^2+bx+c = 0 form (Quadratic)?

Question

What is (x-3)^2 = 5 in ax^2+bx+c = 0 form (Quadratic)

Answer 1

expand x^2-6x+9 = 5
or x^2-6x+4 =0

Answer 2

Say we've 0 = x^2 + 2xy + y To ingredient, only write out 2 contraptions of parenthesis: ( )( ) ascertain the signs and indicators in the midsection: (if the 0.33 variety in the equation is beneficial, the signs and indicators are the two beneficial or the two adverse. if it relatively is adverse, then a million pos. a million neg.) ( + )( + ) What 2 numbers' product equals the 1st variety and may be used to ascertain the 2d variety: (x + )(x + ) What 2 numbers equals the 0.33? (x + y)(x + y) you're executed! If we had x^2 + 6x - 7 = 0 to end the sq.: a million. upload the non variable coefficient to the choice component: x^2 + 6x = 7 2. ingredient out the 'a' term a million(x^2 + 6x) = 7 3. Take a million/2 b, then sq. it and upload it to the two sides a million/2(6) = 3 (x^2 + 6x + (3)^2) = 7 + (3)^2 4. ingredient the quadratic you get and simplify the main appropriate component (x + 3)^2 = sixteen 5. Now, we've 'p' form of a quadratic: (x + 3)^2 = (y + sixteen) The vertex is (-3,-sixteen) 6. resolve for x to get the roots x + 3 = +/- 4 x = -7 x = a million 7. replace x back in to get the y int (to envision, the y int is often the 0.33 term in a quadratic equation) for that reason it relatively is -7 8. There you pass. you have the vertex, axis of sym (while you're graphing up/down parabolas, it relatively is the y term; while you're graphing left/suitable parabolas, it relatively is the x term) the y int/(x int in sideways parabolas) and roots.

Answer 3

It is x^2-6x+4=0.

Answer 4

x^2 - 6x + 4 = 0

Answer 5

yes, this is a quadratic....heres how you go about this problem

you are given (x-3)^2=5

start by foiling the left

(x-3)(x-3)
so you now have
x^2-3x-3x+9=5
or
x^2-6x+9=5
so now you want to move the 5 to the left side because remember we need our equation set equal to zero...to do this subtract 5 from both sides

so you now have
x^2-6x+9-5=0
and your final answer is

x^2-6x+4=0
(here a=1, b=-6, c=4)
hope this helps
matttlocke

Answer 6

x^2-6x+9=5
adding -5
x^2-6x+4=0

Answer 7

(x-3)^2 = 5
x^2-6x+9=5
x^2-6x-4=0