How do we know that the quadratic 3y2−y−12 has real root? (a) Notice the quadratic...





How do we know that the quadratic 3y2y12 has real root?
(a) Notice the quadratic cannot be factored into the product of two binomials with integer cofficients. Does this mean that the quadratic does not have any real roots?
(b) If the answer to part (a) is "no", then explain how we know that the quadratic does have real roots.
(c) Suppose the quadratic has roots y=r and y=s. Find a quadratic with roots r+2and s+2.

Answer & Explanation

Amari Larsen

Amari Larsen


2022-02-02Added 10 answers

(a) This does not mean the quadratic has no real roots. A clearer example of this is
(b) One way to know is to observe that plugging in y=0 and y=3 yields
so somewhere between 0 and 3 the quadratic must equal 0.
Another way to know is by computing the discriminant, which is
The quadratic has a real root because the discriminant is nonnegative.
(c) If r and s are roots of
then it follows that r+2 and s+2 are roots of
which by a little bit of algebra simplifies to
Nevaeh Jensen

Nevaeh Jensen


2022-02-03Added 14 answers

Using Vieta theorem, r+s=13,rs=4.
Thus, (r+2)(s+2)=rs+2(r+s)+4=4+23+4=23
and (r+2)+(s+2)=r+s+4=133
Applying Vieta in reverse, we can get a solution for c):
To show that there is a real root, rewrite the equation as 3(x16)2=12+112 and use that any non-negative real can be written as a square of some real. Or you can use the hint provided and apply Intermediate Value Theorem.

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