SOLUTION: Write a quadratic equation in factored and standard form whose roots are 3 and -4/3. Is there more than one solution possible? Explain!

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Question 1085176: Write a quadratic equation in factored and standard form whose roots are 3 and -4/3. Is there more than one solution possible? Explain!
Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The factors are:
+x+-+3+ and
+x+%2B+4%2F3+
+%28+x+-+3+%29%2A%28+x+%2B+4%2F3+%29+=+0+
Divide the 2nd factor by +3+
+%28+x+-+3+%29%2A%28+3x+%2B+4+%29+=+0++
+3x%5E2+-+9x+%2B+4x+-+12+=+0+
+3x%5E2+-+5x+-+12+=+0+
I can divide by +3+ and get
+x%5E2+-+%285%2F3%29%2Ax+-+12+=+0+
-----------------------------
The equations are:
+y+=+3x%5E2+-+5x+-+12+
and
+y+=+x%5E2+-+%285%2F3%29%2Ax+-+4+
-----------------------------
( -4/3, 0 )
+y+=+3x%5E2+-+5x+-+12+
+0+=+3%2A%28-4%2F3%29%5E2+-+5%2A%28-4%2F3%29+-+12+
+0+=+16%2F3+%2B+20%2F3+-+36%2F3+
+0+=+36%2F3+-+36%2F3+
OK
( 3,0 )
+y+=+3x%5E2+-+5x+-+12+
+0+=+3%2A3%5E2+-+5%2A3+-+12+
+0+=+27+-+15+-+12+
+0+=+27+-+27+
OK
You can check the other form of the equation


Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
When you found a factored polynomial like this

f(x) = %28x-3%29%2A%28x+%2B+4%2F3%29,

you can multiply it by any non-zero constant

F(x) = a%2A%28x-3%29%2A%28x+%2B+4%2F3%29

and you will get another polynomial with the same properties.


Therefore, the solution is not unique.


Actually, there are infinitely many solutions.


They all differ in this multiplier "a" in the factored form.