How do you solve #9x² - 12x + 4 = -3#?

Question

Nathan Axel · Accepted Answer

#x=(2+-sqrt(3)i)/3#

Given #color(white)("XXX")9x^2-12x+4=-3#

There are several ways to solve this. I will demonstrate using a "completing the square method"

The given equation implies #color(white)("XXX")9x^2-12x=-7# (after subtracting #4# from both sides)

#color(white)("XXX")9(x^2-4/3x)=-7#

#color(white)("XXX")9(x^2-4/3x+(2/3)^2)=-7+9 * (2/3)^2#

#color(white)("XXX")9(x-2/3)^2=-7+4#

#color(white)("XXX")(x-2/3)^2=-3/9=-1/3#

#color(white)("XXX")(x-2/3)=+-sqrt(-1/3)#

Note that for Real values #sqrt(-1/3)# is undefined, but if we are allowed Complex values: #color(white)("XXX")x-2/3=+-1/sqrt(3)i# and #color(white)("XXX")x=2/3+-1/sqrt(3)i=(2+-sqrt(3)i)/3#

Why is there not Real solution? Note that the given equation is equivalent to #9x^2-12x+7=0# and here is a graph of #9x^2-12x+7# graph{9x^2-12x+7 [-12.41, 12.9, -4.42, 8.22]} Notice that #9x^2-12x+7# never crosses the X-axis and therefore it is never equal to #0#

Ethan Adkins · Answer

undefined To solve the equation 9x² - 12x + 4 = -3, follow these steps:

1. Add 3 to both sides of the equation to isolate the quadratic term:
9x² - 12x + 4 + 3 = -3 + 3
9x² - 12x + 7 = 0

2. Now, set up the quadratic equation in the form ax² + bx + c = 0:
a = 9, b = -12, and c = 7

3. Use the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / (2a)

4. Substitute the values of a, b, and c into the quadratic formula:
x = (-(-12) ± √((-12)² - 4(9)(7))) / (2(9))

5. Simplify the expression under the square root:
x = (12 ± √(144 - 252)) / 18
x = (12 ± √(-108)) / 18

6. Since the expression under the square root is negative, the equation has complex solutions:
x = (12 ± √(108)i) / 18
x = (12 ± 6√3i) / 18

7. Simplify further by dividing both numerator terms by 6:
x = (2 ± √3i) / 3

Therefore, the solutions to the equation 9x² - 12x + 4 = -3 are x = (2 + √3i) / 3 and x = (2 - √3i) / 3.