A question about quadratics again, sorry!?

Question

I'm working from a text book where the answer to (x-6)^2 = x is given as x=4 or x=9. Also where the answer to 6(x-1)^2 = 5x - 6 is given as x = 3/2 or x = 4/3. I'm not sure how to get to these answers. I know I just asked a similar question but I still need some help! :o) Thanks.

Answer 1

(x-6)^2 = x
x^2 - 12x + 36 = x
x^2 - 12x + 36 - x = 0
x^2 - 13x + 36 = 0
x^2 - (- 4 - 9) + - 4 x - 9 = 0
(x - 4) (x - 9) = 0
If x - 4 = 0
x = 4
If x - 9 = 0
x = 9
x = 4 or x = 9

6(x-1)^2 = 5x - 6
6(x^2 - 2x + 1) = 5x - 6
6x^2 - 12x +6 - 5x + 6 = 0
6x^2 - 17x + 12 = 0
6x^2 - 9x - 8x + 12 = 0
3x(2x - 3) - 4(2x - 3) = 0
(2x - 3)(3x - 4) = 0
If 2x - 3 = 0
2x = 3
x = 3 / 2
If 3x - 4 = 0
3x = 4
x = 4 / 3
x = 3 / 2 or x = 4 / 3

Answer 2

well dear i would help you how you can solve such a question like that;

Part One;

as you know (a - b)^2 = (a^2) + (b^2) -2ab
so we have ; (x-6)^2 = x
(x^2) + 36 - 12x = x
x^2 -12x - x + 36 =
x^2 -13x + 36 ; { here you need to find 2 real numbers like a& b that when you multiply them the result would be +36 & when you add them the result would be -13
(x - a)(x- b) =( x - 4)(x- 9)
x^2 -13x + 36 =( x - 4)(x- 9)
( x - 4)(x- 9) = 0 , so we have
x - 4 = 0 ; x =+4
OR
x - 9 = 0 ; x = +9
_______________

Part Two;

6(x-1)^2 = 5x - 6
{ (x-1)^2 = x^2 +1 - 2x }
6 ( x^2 +1 - 2x ) = 5x - 6
6x^2 +6 -12x = 5x - 6
6x^2 + 6 -12x -5x + 6 =
6x^2 +( -12x - 5x) +(6+6) =
6x^2 - 17x +12 = 0
if ax^2 + bx + c so we have ;
∆ = b^2 - 4ac
if a = +6 , b= -17 & c = +12
∆ = (-17^2) - 4*6*12 =289 - 288 = +1
∆ = +1
x1 =( -b + √ ∆) / 2a =[ -(-17) + √1 ] / 2(6) = +17 +1 / 12 = 18/12 = 3/2
x2 =( -b - √ ∆ ) / 2a = =[ -(-17) - √1 ] / 2(6) = +17 -1 / 12 = 16 /12 = 4/3
x1 = 3/2 & x2 = 4/3

It's waht we are looking for .
Good Luck & very good question.

Answer 3

The general solution for a quadratic equation ax^2+bx+c=0 is
-b+ Sqrt(b^2-4ac) -b- Sqrt(b^2-4ac)
x= ------------------------ and ----------------------
2a 2a

From the given sum, (x-6)^2 = x

x^2 -12x +36 = x
Or, x^2 -13x +36 = 0
where a = 1, b = -13 and c = 36

Substitute the values of a, b, c in the above and get the values of x. But remember, before substituting check the quantity under square root should not be less than 'zero', in which case the solution does not exist.

Good luck.

Answer 4

The link may help you understand quadratics. The name (quadratics) is used because x is squared (not cubed, etc.). If you square anything like 2 x 2 = 4 you get a square (with four sides as in quad for four). Because 2 x 2 = 4 and -2 x -2 = 4 there are usually two answers to "What does x2 (x squared) equal or what is x?" Often only one of the answers makes physical sense and the other answer is discarded. If x2 = 4, x = 2 (or -2). Good luck.

Answer 5

(x - 6)^2 = x
(x - 6)(x - 6) = x
x^2 - 6x - 6x + 36 = x
x^2 - 12x + 36 = x
x^2 - 13x + 36 = 0
(x - 9)(x - 4) = 0
x = 4 or 9

------------------------------------------

6(x - 1)^2 = 5x - 6
6(x - 1)(x - 1) = 5x - 6
6(x^2 - x - x + 1) = 5x - 6
6(x^2 - 2x + 1) = 5x - 6
6x^2 - 12x + 6 = 5x - 6
6x^2 - 17x + 12 = 0
(3x - 2)(4x - 3) = 0

x = (3/2) or (4/3)

When you work the problem down to ax^2 + bx + c form, i would use

x = (-b ± sqrt(b^2 - 4ac))/(2a)

Answer 6

Another easiest way is as below.

Prepare an equation with formula

square of (x)-(sum of values)x+(product of values)=0 and solve. Both should give equal result. For example, here it will be x^2-(4+9)x+4.9=0
which means x^2-13x+36=0. This equation is same as somebody demonstrated above.