^{2 }- 9x + 2 = 0

9 x ^{2 }- 9x + 2 = 0

x=

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Since 9*2=(-3)(-6), and -3-6 = -9,the coefficient of x, you can split -9x into -6x and -3x, and regroup

9x^2 - 9x + 2

= (9x^2 -6x) - (3x-2)

= 3x(3x-2) - (3x-2)

= (3x-2)(3x-1)

Solve (3x-2)(3x-1) = 0, x = 2/3, and x = 1/3

9x^2-9x+2=0

9=3*3 and 2=1*2

use one 3 with 2 and the other 3 with 1

3*2==6 and3*1=3

6+3=9

(3x-2)(3x-1)=0

3x-2=0, 3x=2, x=2/3

3x-1=0, 3x=1, x=1/3

See answer above by Parviz F.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Have to choose 2 numbers in which their product is 18, and their sum is -9, therefore both have to be negative numbers.

ab = 18 a+b = -9

18 = 2 *9 = 6* 3 , so -6-3 = 18 , -6-3 =-9

Split -9X to -6x- 3X

9 X^{2 }- -9x +2 =0

9 X^{2} - 6X - 3X + 2 = 0

3X ( 3X - 2 ) - ( 3X -2 ) = 0 factor by grouping

( 3 X - 2 ) ( 3X -1 ) =0

3X - 2 = 0 **X = 2/ 3** 3X -1 = 0 ** X =**** 1/3**

This is a general rule:

aX^{2 }+ bX + c = 0

choose 2 numbers m,n in such away

that ac = mn and b/a = m+n

where you have the values of a,b,c given in a quadratic equation.

If m.n are integer values, then you can factor like the above example, otherwise use quadratic formula:

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