How do you solve #0=10x^2 + 9x-1# using the quadratic formula?

Question

Genesis Allen · Accepted Answer

I will show you how and let you do the calculations

First you need to know the standard form which is: #y=ax^2+bx+c# In your case set: #y=0# This is the condition needed any way to find where the graph crosses the x-axis. #a=10# #b=9# #c=(-1)# the negative or minus is very important These are then substituted into: #x = ( b +- sqrt( b^2 - 4ac))/(2a)# If the curve crosses the line in two places you will have two answers. If it is such that the x-axis is tangential to the curve then you only have one solution. If the curve is such that it does not cross the x-axis then (I think!) # b^2 - 4ac# is negative. You will need to check it!! In your case you will have: #(9 +- sqrt( 9^2 - (4)(10)(-1)))/(2 times 10)# Notice that I use brackets to make sure that the positive or negative state of the values can be included. Reduces confusion!!! Now you have a go at doing the calculation!!!

Ethan Adkins · Answer

undefined To solve the equation 0 = 10x^2 + 9x - 1 using the quadratic formula, where the general form of a quadratic equation is ax^2 + bx + c = 0, follow these steps:

1. Identify the coefficients a, b, and c:  
   - a = 10
   - b = 9
   - c = -1

2. Substitute the coefficients into the quadratic formula:  
   x = (-b ± √(b^2 - 4ac)) / (2a)

3. Plug in the values for a, b, and c into the quadratic formula:  
   x = (-(9) ± √((9)^2 - 4(10)(-1))) / (2(10))

4. Simplify the expression inside the square root:  
   b^2 - 4ac = (9)^2 - 4(10)(-1) = 81 + 40 = 121

5. Substitute the simplified expression back into the quadratic formula:  
   x = (-(9) ± √(121)) / (2(10))

6. Find the square root of 121:  
   √(121) = 11

7. Substitute the square root value into the equation:  
   x = (-(9) ± 11) / 20

8. Now, solve for both possible values of x:  
   x₁ = (-(9) + 11) / 20 = 2 / 20 = 1/10  
   x₂ = (-(9) - 11) / 20 = -20 / 20 = -1

Therefore, the solutions to the equation 0 = 10x^2 + 9x - 1 are x = 1/10 and x = -1.