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Re: f(x)=3x^2−15x+16 [#permalink]
If i factorize f(x)=3x2−15x+16 , getting b^2-4ac = √33 .means solution will not give real values
but applying formula , it seems we have real solution −b/a= 5

can someone explain?
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Re: f(x)=3x^2−15x+16 [#permalink]
1
greMS15 wrote:
If i factorize f(x)=3x2−15x+16 , getting b^2-4ac = √33 .means solution will not give real values
but applying formula , it seems we have real solution −b/a= 5

can someone explain?



Can you plz check the link https://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php

and identify where u went wrong.
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Re: f(x)=3x^2−15x+16 [#permalink]
I do not know what i have to look for in this link

pranab01 wrote:
greMS15 wrote:
If i factorize f(x)=3x2−15x+16 , getting b^2-4ac = √33 .means solution will not give real values
but applying formula , it seems we have real solution −b/a= 5

can someone explain?



Can you plz check the link https://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php

and identify where u went wrong.
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Re: f(x)=3x^2−15x+16 [#permalink]
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Expert Reply
This

Attachment:
shot26.jpg
shot26.jpg [ 145.02 KiB | Viewed 3046 times ]


it clears show to you how to solve a quadratic equations
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Re: f(x)=3x^2−15x+16 [#permalink]
probably i am missing something obvious?

i am not able to factorize the equation 3x2−15x+16 :), i am getting solutions that is not real. b^2-4ac i am getting 33 , is that correct?



Carcass wrote:
This

Attachment:
shot26.jpg


it clears show to you how to solve a quadratic equations
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Re: f(x)=3x^2−15x+16 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: f(x)=3x^2−15x+16 [#permalink]
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