[MOVED] Sum of roots of kx^2-(1+k)x+(3k+2)=0 twice product of roots. Find "k", roots.

[Nikolas111]

[MOVED] Sum of roots of kx^2-(1+k)x+(3k+2)=0 twice product of roots. Find "k", roots.

The quadratic equation : kx^2-(1+k)x+(3k+2)=0 is such that the sum of its roots is twice their product.Find k and the two roots.

Could somebody help me please with this quadratic equation the answers of this equation are k=-3/5 and roots = -1 and 1/3 I would need to know the way how it was counted. Thank you in advance

 

[stapel]

The quadratic equation : kx^2-(1+k)x+(3k+2)=0 is such that the sum of its roots is twice their product.Find k and the two roots.

Could somebody help me please with this quadratic equation the answers of this equation are k=-3/5 and roots = -1 and 1/3 I would need to know the way how it was counted.

You can review similar questions, such as this or this. Then use what you know about algebra and quadratics.

You know the Quadratic Formula (here), so you can get started by plugging into that. What are the expressions for the two roots, in terms of "k"?

You are given that the sum of the roots is twice the product. What expression stands for "the sum of the roots"? What mathematical symbol stands for "is"? What expression stands for "the product of the roots"? What expression stands for "twice the value of the product of the roots"? What equation can you create from this? When you solve that equation, what value(s) do you get for "k"? And so forth.

If you get stuck, please reply showing your thoughts and efforts so far. Thank you!

 

[pka]

The quadratic equation : kx^2-(1+k)x+(3k+2)=0 is such that the sum of its roots is twice their product.Find k and the two roots.

Could somebody help me please with this quadratic equation the answers of this equation are k=-3/5 and roots = -1 and 1/3 I would need to know the way how it was counted. Thank you in advance

In a quadratic, \(\displaystyle ax^2+bx+c=0\) the sum of the roots is \(\displaystyle \dfrac{-b}{a}\) and the product is \(\displaystyle \dfrac{c}{a}\).

 

[Nikolas111]

A quadratic polynomial equation

Actually, I don't know if I started correctly, my first step was that I removed brackets I got kx^2-x-kx +3k+2=0 then I factorized it by k*(x^2-x+3) +2-x=0 but in my opinion it wasn't the right way because I always got a wrong answer .

 

[Harry_the_cat]

\(\displaystyle kx^2-(1+k)x+(3k+2) =0\)

Here \(\displaystyle a = k\), \(\displaystyle b=-(1+k)\), \(\displaystyle c=(3k+2)\)

As pka said "In a quadratic, \(\displaystyle ax^2 + bx +c =0\) the sum of the roots is \(\displaystyle \frac{-b}{a}\) and the product of the roots is \(\displaystyle \frac{c}{a}\)."

In your question, the sum of the roots is twice the product of the roots.

This means that \(\displaystyle \frac{-b}{a} = \frac{2c}{a} \)

Now combine these two ideas and solve for k.

 

[Nikolas111]

A quadratic polynomial equation

Thank you very much for helping me