Roots of a quadratic equation.b

**I would welcome your advice on my workings in this question with particular reference to part (ii).**

Thanking you in advance.

The roots of the quadratic equation$\displaystyle x_2 + (x + 1) = k$ are $\displaystyle {\alpha}+{\beta}$

(i) prove that $\displaystyle {\alpha}_2 + {\beta}_2 = k$;

$\displaystyle 2x_2 + 2x + 1 - k = 0$

$\displaystyle {\alpha} + {\beta} =\frac{-b}{a}= -1$; $\displaystyle {\alpha}{\beta} = \frac{c}{a} = \frac{1-k}{2}$

$\displaystyle {\alpha}_2 + {\beta}_2 = ({\alpha} + {\beta})_2 - 2{\alpha}{\beta} = (-1)_2 + 2(\frac{1 - k}{2}) = -1 + 1 + k = k$; Hence $\displaystyle {\alpha}_2 + {\beta}_2 = k$

(ii) Find, in terms of $\displaystyle k$, a quadratic equation in $\displaystyle x$, whose roots are $\displaystyle {\alpha}_2$ and $\displaystyle {\beta}_2$.

Formula for quadratic equation where $\displaystyle {\alpha}$ and $\displaystyle {\beta}$ are the roots of the equation $\displaystyle ax_2 + bx + c = 0$:

$\displaystyle x_2$ - sum of roots + product of roots;

Where the equation's roots are $\displaystyle {\alpha}_2$ and $\displaystyle {\beta}_2$: we know from (i) above that $\displaystyle {\alpha}_2 + {\beta}_2 = k$ and that $\displaystyle {\alpha}{\beta} = \frac{1 - k}{2}$;

so $\displaystyle x_2$ - sum of roots + product of roots = 0 implies $\displaystyle 2x_2 - 2kx + (\frac{1 - k}{2})_2$

According to the textbook the answer is $\displaystyle 4x_2 - 4kx + (1 - k)_2 = 0$

I might have erred earlier on, but from my workings shouldn't the answer be $\displaystyle 8x_2 - 8kx + (1 - k)_2$

in other words, one cannot multiply by 2 across the equation as that 2 is part of a fraction that should be squared.

**Thank you for taking the time to read this post.**

Re: Roots of a quadratic equation.b

Quote:

Originally Posted by

**Seaniboy** The roots of the quadratic equation$\displaystyle x_2 + (x + 1) = k$ are $\displaystyle {\alpha}+{\beta}$

(i) prove that $\displaystyle {\alpha}_2 + {\beta}_2 = k$;

$\displaystyle 2x_2 + 2x + 1 - k = 0$

$\displaystyle {\alpha} + {\beta} =\frac{-b}{a}= -1$; $\displaystyle {\alpha}{\beta} = \frac{c}{a} = \frac{1-k}{2}$

$\displaystyle {\alpha}_2 + {\beta}_2 = ({\alpha} + {\beta})_2 - 2{\alpha}{\beta} = (-1)_2 + 2(\frac{1 - k}{2}) = -1 + 1 + k = k$; Hence $\displaystyle {\alpha}_2 + {\beta}_2 = k$

(ii) Find, in terms of $\displaystyle k$, a quadratic equation in $\displaystyle x$, whose roots are $\displaystyle {\alpha}_2$ and $\displaystyle {\beta}_2$.

Formula for quadratic equation where $\displaystyle {\alpha}$ and $\displaystyle {\beta}$ are the roots of the equation $\displaystyle ax_2 + bx + c = 0$:

$\displaystyle x_2$ - sum of roots + product of roots;

Where the equation's roots are $\displaystyle {\alpha}_2$ and $\displaystyle {\beta}_2$: we know from (i) above that $\displaystyle {\alpha}_2 + {\beta}_2 = k$ and that $\displaystyle {\alpha}{\beta} = \frac{1 - k}{2}$;

so $\displaystyle x_2$ - sum of roots + product of roots = 0 implies $\displaystyle 2x_2 - 2kx + (\frac{1 - k}{2})_2$

According to the textbook the answer is $\displaystyle 4x_2 - 4kx + (1 - k)_2 = 0$

Does anyone else find the posting incomprehensible? Or is just me?

**Quadratic equation** ? Where? Are some of the subscripts suppose to be exponents? Why subscript the $\displaystyle \alpha~\&~\beta~?$

Re: Roots of a quadratic equation.b

(sighs)

I spent about 10 minutes looking at this and after posting several versions of different things that would make the post make more sense and all I can come up with is that Plato's right. This is sheer gibberish.

@seaniboy You are going to have to repost this after a lot of editing. Start with Plato's suggestions.

-Dan

Re: Roots of a quadratic equation.b

The subscript 2's are meant to be squared signs. I don't know how to go about doing the question though.

Seaniboy in Latex the symbol ^ is used for making sqaure signs, x^2 is x sqaured

Re: Roots of a quadratic equation.b

Quote:

Originally Posted by

**Shakarri** The subscript 2's are meant to be squared signs. I don't know how to go about doing the question though.

Seaniboy in Latex the symbol ^ is used for making sqaure signs, x^2 is x sqaured

@Shakarri, you are not Seaniboy . So why do you think that are we are obligated to guess as to what the OP actually means?

Are you sure that is what it means? If so, then thank you. If you are not sure then stay out of it.

Re: Roots of a quadratic equation.b

Yes I am sure, every line makes sense if the subscript is a superscript. Part (i) is a pretty standard question on the pre-university Irish course.

Re: Roots of a quadratic equation.b

Apologies. It's the first time that I used latex. I will try and repost in a more comprehensive, less turgid format. But thank you for taking the time to reply.

Re: Roots of a quadratic equation.b

Thank you @Shakarri and @Plato. I need to take a crash course in Latex and not maths. Thank you both for taking the time to reply.

Re: Roots of a quadratic equation.b

I will, Dan. Thank you for your input and apologies for the mess.

Sean.

Re: Roots of a quadratic equation.b

OK folks - here goes again...

QUESTION: FROM THE EQUATION $\displaystyle x^2 + (x + 1)^2 = k$

FIND IN TERMS OF K, A QUADRATIC EQUATION WHOSE ROOTS ARE ALPHA SQUARED AND BETA SQUARED

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Re: Roots of a quadratic equation.b

Re: Roots of a quadratic equation.b

Thank you for that explanation. Following your workings, we get the equation:

1. $\displaystyle 2x^2 - 2kx + (\frac{1 - k}{2})^2 = 0$; however, the answer in the textbook is:

2. $\displaystyle 4x^2 - 4kx + (1-k)^2 = 0$, which is multiplying the coefficient of each term by 2.

QUESTION: given that 2 is part of a fraction in the last term of the equation that should be squared - see 1. above, why are the coefficients not multiplied by 4 instead?

Apologies for being awkward and thank you once again for explaining so clearly.

Sean.

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Re: Roots of a quadratic equation.b

It is the same answer what we got. see the steps further. Attachment 28798

Re: Roots of a quadratic equation.b

Thanks. I understand now. Really appreciate you taking the time to help me.

Cheers, Sean.