Find the values of k that make 2x2 + kx + 5 factorable.

7 or 11
2 or 5
1 or 10
1 or 5

Question

Find the values of k that make 2x2 + kx + 5 factorable.


       7 or 11
       2 or 5
       1 or 10
       1 or 5

anonymous · Answer

I was thinking A

solomonzelman · Answer

yes

solomonzelman · Answer

A is right

solomonzelman · Answer

you plugged the options in, correct ?

anonymous · Answer

yes I did do you want to see my work?

solomonzelman · Answer

I believe you, and you got it correctly:)`

anonymous · Answer

thank you ! :)

solomonzelman · Answer

I was going to post a prove:
$\large\color{black}{ \displaystyle k^2-4(2)(5)  }$
$\large\color{black}{ \displaystyle k^2-4(10)  }$
$\large\color{black}{ \displaystyle k^2-40  }$

solomonzelman · Answer

I am doing a discriminant test, and you know that for it to be factor-able, the discriminant has to be a perfect square (like 1, 4, 9, 16, 25 and on...).
And certainly it can't be negative,

Therefore  $\large\color{black}{ \displaystyle k^2\ge40  }$

solomonzelman · Answer

And ergo, A is the correct answer, because in other options you have at least 1 value (for k) that doesn't satisfy the statement $\large\color{black}{ \displaystyle k^2\ge40  }$.

solomonzelman · Answer

hope this logic makes sense...

solomonzelman · Answer

yw, tho

anonymous · Answer

oh ok I see how you got it 

 the discriminant of a function ax^2 + bx + c is b^2 - 4ac so... ur prob 2x^2 + kx + 5 ax ^2 + bx + c the discriminant is.. b^2 - 4ac k^2 -40 and this value should be above 0 which means.. k^2 - 40 > 0 k^2> 0 and k^2 - 40 should be a perfect square.... the choices that satisfy this conditions r.. 7 , 11

solomonzelman · Answer

well, i didn't say it should be above zero, as (although 0 is not considered a perfect square, as regards to this discriminant test) it does only give you that the trinomial is a perfect square trinomial.

$\large\color{black}{ \displaystyle k^2\ge40  }$
and not $\large\color{red}{ \displaystyle k^2>40  }$

BUT, you got the concept

solomonzelman · Answer

well, the bottom line is...

solomonzelman · Answer

$\normalsize\color{royalblue}{     \rm Discriminant  ~=~perfect~square      }$   the trinomial can be factored into integer roots. (perfect square implies that discriminant is AT LEAST 0)

$\normalsize\color{royalblue}{     \rm Discriminant  ~>~0      }$    some real number solutions

$\normalsize\color{royalblue}{     \rm Discriminant  ~=~0      }$   trinomial is a perfect square 

$\normalsize\color{royalblue}{     \rm Discriminant  ~<~0      }$   some imaginary solutions