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son4ous [18]
3 years ago
14

◆ Quadratic Equations ◆Please help !

Mathematics
1 answer:
Mila [183]3 years ago
3 0
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=\dfrac{-b\pm\sqrt{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=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}

Solve for 'x':

\sf x=\dfrac{4\pm\sqrt{16-8}}{2}

\sf x=\dfrac{4\pm\sqrt{8}}{2}

\sf x\approx\dfrac{4\pm 2.83}{2}

\sf x\approx 0.59,3.41

So the answer would be A.
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