How do you use the quadratic formula to solve for the roots in the following equation #4x^2+5x+2=2x^2+7x-1#?

Question

Nathan Axel · Accepted Answer

#x=(1+2i)/2# or #(1-2i)/2# #->4x^2-2x^2+5x-7x+2-(-1)=0# #->2x^2-2x+3=0# And then you use the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)# for the equation #ax^2+bx+c=0# and here we have got #a=2#, #b=-2# and #c=3# hence #x=(-(-2)+-sqrt((-2)^2-4xx2xx3))/(2xx2)# = #(2+-sqrt(-16))/4=(2+-4i)/4# i.e. #x=(1+2i)/2# or #(1-2i)/2#

Abigail Allen · Answer

undefined To solve the equation 4x^2 + 5x + 2 = 2x^2 + 7x - 1 using the quadratic formula, follow these steps:

1. Rearrange the equation to set it equal to zero: 2x^2 + 7x - 1 - (4x^2 + 5x + 2) = 0.
2. Combine like terms to simplify: (2x^2 + 7x - 1) - (4x^2 + 5x + 2) = 0 becomes -2x^2 + 2x - 3 = 0.
3. Identify the coefficients: a = -2, b = 2, and c = -3.
4. Apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
5. Substitute the coefficients into the quadratic formula: x = (-2 ± √(2^2 - 4*(-2)*(-3))) / (2*(-2)).
6. Simplify the expression under the square root: x = (-2 ± √(4 - 24)) / (-4).
7. Continue simplifying: x = (-2 ± √(-20)) / (-4).
8. Since the expression under the square root is negative, the roots will involve complex numbers.
9. Simplify further: x = (-2 ± 2i√5) / (-4).
10. Divide both the numerator and the denominator by -2: x = (1 ± i√5) / 2.
11. The roots are thus x = (1 + i√5) / 2 and x = (1 - i√5) / 2.