SOLUTION: One of the five quadratics below has a repeated root. (The other four have distinct roots.) What is the repeated root? -x^2 + 18x + 81 3x^2 - 6x + 9 8x^2 - 32x + 32 25x^2 -

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: One of the five quadratics below has a repeated root. (The other four have distinct roots.) What is the repeated root? -x^2 + 18x + 81 3x^2 - 6x + 9 8x^2 - 32x + 32 25x^2 -      Log On


   

Question 1209117: One of the five quadratics below has a repeated root. (The other four have distinct roots.) What is the repeated root?
-x^2 + 18x + 81
3x^2 - 6x + 9
8x^2 - 32x + 32
25x^2 - 30x - 9
x^2 - 14x + 196
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

The quadratic template is y = ax^2+bx+c
The discriminant formula is d = b^2-4ac.
This is the stuff under the square root in the quadratic formula.

Here are the possible cases
  • d > 0 yields 2 distinct roots
  • d = 0 gives repeated roots
  • d < 0 means there aren't any real number solutions. The two solutions are complex.
Let's find the discriminant of -x^2 + 18x + 81
d = b^2-4ac
d = 18^2-4*(-1)*81
d = 648
The result is positive, so there will be 2 distinct roots.
Therefore we can rule out -x^2 + 18x + 81 as a possible answer.

Now let's try 3x^2 - 6x + 9
d = b^2-4ac
d = (-6)^2-4*3*9
d = -72
The negative discriminant indicates there will be two complex solutions of the form a+bi where i = sqrt(-1).
We can rule out 3x^2 - 6x + 9 as a possible answer.

I'll let the student finish up with the remaining 3 quadratics to test out.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
One of the five quadratics below has a repeated root. (The other four have distinct roots.) What is the repeated root? 
    (1)  -x^2 + 18x + 81
    (2)  3x^2 - 6x + 9
    (3)  8x^2 - 32x + 32
    (4)  25x^2 - 30x - 9
    (5)  x^2 - 14x + 196
~~~~~~~~~~~~~~~~~~~~~~ 

If a quadratic polynomial has repeated root, then the polynomial is

    +/- (ax-b)^2,  or  +/- (a^2 - 2abx + b^2).


We see that if a quadratic polynomial has repeated root, then the coefficient at x^2 and the constant term
have the same sign: they either both are positive or both are negative,


So, cases (1) and (4) fall out;  cases (2), (3) and (5) remain for the further consideration.


In case (5),  the coefficient at x should be  +/- 2*sqrt(coef at x^2)*sqrt(constant term) = +/- 2*sqrt(1)*sqrt(196) = +/- 28,
              but we have there number -14 instead, so case (5) falls out.  Cases (2) and (3) remain for further consideration.


In case (3) ,  8x^2 - 32x + 32 = 8(x^2 -4x + 4) = 8*(x-2)^2  has the repeated root 2,
               so, it is that unique case, we are searching for.


At this point, the unique case is just found; so, case (2) does not require further analysis.


ANSWER.  The unique polynomial with repeated root is case (3).

         The repeated root is 2.

Solved.