Question 1060712: Solve the equation by completing the square:
5x^2-9x+4=0 Found 3 solutions by josgarithmetic, addingup, MathTherapy:Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website! Completing the Square to Solve a Quadratic Equation, LESSON Answer by addingup(3677) (Show Source): You can put this solution on YOUR website! To complete the square we have to get the equation into the proper form, which is: x^2+bx = c
:
5x^2-9x+4 = 0
5x^2+(-9)x = -4
[5x^2+(-9)x]/5 = -4/5
x^2+(-9/5)x = -4/5
now take the coefficient of x, which is -9/5, and divide it by 2. All you have to do is multiply the divisor, 5, times 2 and you've automatically divided -9/5 by 2: -9/5 divided by 2 becomes -9/10. Square this number (when you square a negative it becomes positive because -*- = +): 81/100 Now add this number to both sides of the equation:
x^2+(-9/10)x+81/100 = -4/5+81/100 add on the right (find a common denominator and add the numerators):
x^2+(-9/10)x+81/100 = 1/100
factor the left side:
(x-9/10)^2 = 1/100
Take the square root of both sides and you get:
x-9/10 = 1/10 or x-9/10 = -1/10
add 9/10 on both sides of both equations
x = 1 or x = 8/10 = 4/5 = 0.8 if you need it in decimals.
:
John
------ Moving the constant or c to the right side ------ Dividing entire equation by 5 to get a leading coefficient of + 1 ------ Taking of b, squaring it, and adding the result to both sides of equation ----- Taking the square root of both sides