Question

◆ Quadratic Equations ◆
Please help !

Answer 1

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4


`\sf ax^2+(1-a(b+c))x+abc-d)`


`\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))`

Simplify into standard form:


`\sf x^2+(1-1(5))x+6-4`


`\sf x^2+(1-5)x+2`


`\sf x^2-4x+2`

Use the quadratic formula to solve:


`\sf x=(-b\pm√(b^2-4ac))/(2a)`

For functions in the form of
`\sf ax^2+bx+c`. So in this case:

a = 1
b = -4
c = 2

Plug them in:


`\sf x=(4\pm√((-4)^2-4(1)(2)))/(2(1))`

Solve for 'x':


`\sf x=(4\pm√(16-8))/(2)`


`\sf x=(4\pm√(8))/(2)`


`\sf x\approx(4\pm 2.83)/(2)`


`\sf x\approx 0.59,3.41`

So the answer would be A.