How do we know that the quadratic 3y2−y−12 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.

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How do we know that the quadratic 3y2−y−12 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 &quot;no&quot;, 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.

Amari Larsen · Accepted Answer

(a) This does not mean the quadratic has no real roots. A clearer example of this is  y2−2=(y+2)(y−2).  (b) One way to know is to observe that plugging in y=0 and y=3 yields  3⋅02−0−12=−12,  3⋅32−3−12=12.  so somewhere between 0 and 3 the quadratic must equal 0.   Another way to know is by computing the discriminant, which is  Δ=b2−4ac=(−1)2−4⋅3⋅(−12)=145.  The quadratic has a real root because the discriminant is nonnegative.   (c) If r and s are roots of  3y2−y−12,  then it follows that r+2 and s+2 are roots of  3(y−2)2−(y−2)−12  which by a little bit of algebra simplifies to  3y2−13y+2.

Nevaeh Jensen · Accepted Answer

Using Vieta theorem, r+s=13,r⋅s=−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):  3y2−13y+2=0  To show that there is a real root, rewrite the equation as 3(x−16)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.

How do we know that the quadratic 3y^{2}-y-12 has real root? (a) Notice the quadratic cann

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