Question

One function, f(x), with two real rational solutions.
Can someone help me make onit.

Answer 1

Pick any two rational numbers as roots. For example 2 and 3.
If 2 is a root of a quadratic equation then (x-2) will be a factor of the quadratic equation.
If 3 is a root of a quadratic equation then (x-3) will be a factor of the quadratic equation.

So the quadratic equation is: (x - 2)(x - 3) = 0

Multiply the left hand side and simplify.

Answer 2

Also, for any equation \[Ax ^{2}+Bx + C = 0\]
with non negative coefficients A, B, and C
there are two real rational solutions given by
\[x= (- B ^{2} \pm \sqrt{B ^{2} - 4AC}) \div (2A)\]

Answer 3

Hmm ok so i came up with x^2 + 16x + 10 ...so I can have my two rational numbers because my 10 has two numbers that square to it

Answer 4

I never really been taught that yet @Matt.Mawson

Answer 5

@ranga actually idk if I did that right cause when I did b^2 - 4ac i get a negative number

Answer 6

Following up on my example of having picked 2 and 3 as roots, the quadratic equation is: (x - 2)(x - 3) = 0
x^2 - 3x - 2x + 6 = 0
x^2 - 5x + 6 = 0

Answer 7

Alrighty!

Answer 8

okay.

Answer 9

so do the b2 - 4ac

Answer 10

x^2 - 5x + 6
a = 1, b = -5, c = 6
b^2 - 4ac = (-5)^2 - (4)(1)(6) = 25 - 24 = 1

Answer 11

Oh wow. I forgot to square. Thanks so much!

Answer 12

u r welcome.