Help ! Quadratic Equations.?

Question

Show that the quadratic equation 3x^2 + p = (p+3)x has real roots for all values of p.

Show me every step please.

Techno, p could be non-real numbers, couldn't it?

Alright, ignore the "details" before this statement.

Answer 1

3x² + p = (p + 3)x



We know that it is a quadratic equation, so it is always useful to transpose all terms to the left...

3x² + p - (p + 3)x = 0



Then, we rearrange the terms so that the x powers are in descending order (i.e. x³, x², x etc.)

3x² - (p + 3)x + p = 0



Now to know what are the limits of the values of p, we use the discriminant formula d = b² - 4ac. if d ≥ 0, then the equation has real roots, but if d < 0, then it has no real roots. Thus,

d = [-(p + 3)]² - 4(3)(p)



And

d = p² + 6p + 9 - 12p



And,

d = p² - 6p + 9



We may again factor,

d = (p - 3)²



Now since p is a real number, p - 3 is also a real number and the square of a real number is never negative. So,
(p - 3)² ≥ 0

Therefore, by substitution
d ≥ 0

Which tells us that the quadratic equation has real roots for all values of p.

^_^

Answer 2

First of all rearrange the terms:

3x^2 - (p+3)x + p = 0

then use this equation to find the roots:

x1 = (-b+sqrt(b^2) + 4ac))/2a = ((p+3)+sqrt((p+3)^2 + 4(3)(p))/6

X2 = (-b+sqrt(b^2) + 4ac))/2a = ((p+3)-sqrt((p+3)^2 + 4(3)(p))/6

sqrt = square root

which implies that what ever value of p is we have real roots

Answer 3

easy
take (p+3) to the left
3x^2-(p+3)x + p =0
or 3x^2-3x -px + p =0

or 3x(x-1)-p(x-1) = 0
or(3x-p)(x-1) = 0
x = 1 or p/3 real for all p in real