Quadratics question in Mathematics?

9(x^2 - 4x + 52/9)

Half the coefficient of x (-4) to get -2. Square -2 to get 4. Next, write +4 - 4

9(x^2 - 4x + 4 - 4 + 52/9)...........You've just completed the squares

9[(x - 2)^2 + 16/9]...........Note that (x - 2)^2 = x^2 - 4x + 4

Open brackets

9(x - 2)^2 + 16

Hence, or otherwise, find the set of values taken by 9x² - 36x + 52 for real x ???

If you equate 9x² - 36x + 52 to zero,

9(x - 2)^2 + 16 = 0

(x - 2)^2 = -16/9

x - 2 = +/-i*4/3.........'i' is a complex number and i^2 = -1

x = 2+/-4i/3

When 9x² - 36x + 52 is equated to zero, there'll be no real value for x

Alternatively in Ax^2 + Bx + C = 0, x will be a complex root if B^2 is less than 4AC

For 9x^2 - 36x + 52 = 0,

(-36)^2 is less than 4*9*52..........Hence the complex value of x

9x² - 36x + 52

We will need to divide by 9 to get 9(x² - 4x+ 52/9). then complete the square

by multiplying - 4 by 1/2 which gives -2. Now square -2 which is 4

Now add 4 to the equation and subtract 4 from the same equation as well. In a way we are adding zero to the equation which doesn't change the equation.

9(x² - 4x+ 4 -4+ 52/9)=9(x² - 4x+ 4+ (- 4+ 52/9)) = 9(x² - 4x+ 4+ (- 36+ 52)/9)

Now x² - 4x+ 4 = (x-2)² and (- 36+ 52)/9 = 16/9, thus we get

9((x-2)² + 16/9) Distribute the 9 to get

9(x-2)² + 16

(3x-6)² + 16 By comparing A = 3, B= 6 and C= 16 vertex is ( 6,16)

by setting the final results to zero we obtain

9(x-2)² + 16=0

9(x-2)² = -16

(x-2)² = -16/9

(x-2)² = i²16/9 since i² = - 1

x-2 =±√ (i²16/9)

x = 2 ±√ (i²16/9)

x = 2 ±i 4/3 Hence complex roots

(Ax - B)² + C = A²x² - 2ABx + B² + C = 9x² - 36x + 52

A²x² = 9x²; A = 3

-2ABx = -36x; B = 6

B² + C = 52; C = 16

9x² - 36x + 52 = (3x - 6)² + 16

If x = 2, (3x - 6)² + 16 = 16

For all other values of x, (3x - 6)² is positive and (3x - 6)² + 16 > 16

Hence 16 < (9x² - 36x + 52) < ∞ for all real x

9x^2-36x+52

(3x)^2-2*3x*6+6^2-6^2+52

(3x-6)^2 +52-36

(3x-6)^2 +16

A=3

B=6

C=16

The lowest value of (3x-6)^2 is 0 as square of a number can't be negative and highest value tends to infinity

So for x belongs to real numbers y value lies between [16, infinity)

9x² - 36x + 52 = (Ax - B)² + C

9x² - 36x + 52 = A²x² - 2ABx + B² + C

9x² = A²x²

A = ±3

-36x = -2ABx

AB = 18

B = 18/A = ±6

B² + C = 52

C = 16

9x² - 36x + 52 = (3x - 6)² + 16

also,

9x² - 36x + 52 = (-3x + 6)² + 16

y = 9(x^2 - 4x + 4) + 52 - 36

= 9(x - 2)^2 + 16