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

treslagosnv

treslagosnv

Answered

2022-02-01

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

Expert

2022-02-02Added 10 answers

(a) This does not mean the quadratic has no real roots. A clearer example of this is
y22=(y+2)(y2).
(b) One way to know is to observe that plugging in y=0 and y=3 yields
302012=12,
332312=12.
so somewhere between 0 and 3 the quadratic must equal 0.
Another way to know is by computing the discriminant, which is
Δ=b24ac=(1)243(12)=145.
The quadratic has a real root because the discriminant is nonnegative.
(c) If r and s are roots of
3y2y12,
then it follows that r+2 and s+2 are roots of
3(y2)2(y2)12
which by a little bit of algebra simplifies to
3y213y+2.
Nevaeh Jensen

Nevaeh Jensen

Expert

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):
3y213y+2=0
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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