How do you solve #5x^2+3x-4=0# by completing the square?

Question

Genesis Allen · Accepted Answer

The roots are

#x_1=(-3+sqrt(89))/10#
#x_2=(-3-sqrt(89))/10# Using the provided #5x^2+3x-4=0# Subtract five from the initial two terms. #5(x^2+3/5x)-4=0# Now take the number 3/5, divide it by 2, and square the result to get 9/100. The terms inside the grouping symbol will then be added to and subtracted from this value. Let's proceed. #5(x^2+3/5x)-4=0# Five(x^2+3/5x+9/100-9/100)-4=0# Our perfect square trinomial is now available. In order for #x^2+3/5x+9/100=(x+3/10)^2# Five((x+3/10)^2-9/100)-4=0# Now reverse the -4 to the right side of the equation, divide both sides by 5, and then reverse the -9/100 to the right as well. (x+3/10)^2-9/100) = #5# #4/5# #(cancel5((x+3/10)^2-9/100))/cancel5 #4/5#= #(x+3/10)^2-9/100= #2=4/5+9/100# #(x+3/10)^2 streamline #(x+3/10)^2 = (80+9) / 100# #(x+3/10)^2=89/100# Determine both sides' square roots. (x+3/10)^2) #sqrt=+-sqrt(89/100)# +-sqrt(89/100)# = #x+3/10

Nathan Axel · Answer

undefined To solve the quadratic equation 5x^2 + 3x - 4 = 0 by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   5x^2 + 3x = 4

2. Divide all terms by the coefficient of x^2 to make the leading coefficient 1:
   x^2 + (3/5)x = 4/5

3. Take half of the coefficient of x, square it, and add it to both sides of the equation to complete the square:
   x^2 + (3/5)x + (3/10)^2 = 4/5 + (3/10)^2
   x^2 + (3/5)x + 9/100 = 4/5 + 9/100

4. Simplify both sides:
   x^2 + (3/5)x + 9/100 = 80/100 + 9/100
   x^2 + (3/5)x + 9/100 = 89/100

5. Rewrite the left side as a squared binomial:
   (x + 3/10)^2 = 89/100

6. Take the square root of both sides:
   x + 3/10 = ±√(89/100)

7. Solve for x:
   x = -3/10 ± √(89/100)

8. Simplify the expression:
   x = -3/10 ± √89/10

Thus, the solutions are x = (-3 + √89)/10 and x = (-3 - √89)/10.