# Solving quadratic equation $x^2-(k+1)x+k+1=0$ using quadratic formula

## Disclaimer

I am new here so please mind how i have executed this question. Feel free to comment on how to improve as i have read on “how to ask a good question”.

## Problem

I am having trouble answering this quadratic equation that my teacher at school used. I do not know where she had gotten this equation from.

Equation:
$$x^2-(k+1)x+k+1=0$$

Quadratic Formula: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

My teacher got the final answer of $$x=\frac{(k+1)\pm k}{2}$$

I have asked my teacher on how to solve for x, however she said she won’t show the steps on how to solve this problem because it is simply too easy. When i don’t understand how she got that answer.

## My working out

I have simply found my coefficients for x^2 and x, and my constant is k+1.

a= 1
b= (-k-1)
c= k+1

I am not sure if c actually equals k+1
Then I put them into the quadratic formula.

$$x=\frac{-(-k-1)\pm\sqrt{(-k-1)^2-4\times 1\times k+1}}{2\times 1}$$

Which then i get

$$x=\frac{(k+1)\pm\sqrt{k^2+1-4k-3}}{2}$$

• Your teacher's answer is wrong and you have also made mistake as pointed out in Toby Mak's answer. – Kavi Rama Murthy Apr 19 '19 at 9:58
• If the teacher was right then $c$ must equal $1/2\cdot(k+1/2)$ due to Vieta. – Michael Hoppe Apr 19 '19 at 15:14

You forgot that $$(-k-1)^2 \ne (-k)^2 + (-1)^2$$. Where's the middle term?

$$x=\frac{-(-k-1)\pm\sqrt{k^2+2k+1-\color{red}{4(k+1)}}}{2\times 1}$$ $$x=\frac{k+1\pm\sqrt{k^2-2k-3}}{2}$$

which does not simplify to your teacher's answer (as others have described).

• Thank you so much. I appreciate your help! – Sophia Serrato Apr 19 '19 at 10:02
• Just another thing: you can click the tick (below the voting bar) to accept an answer. You don't have to accept now as there may be more answers in the future. – Toby Mak Apr 19 '19 at 10:05

Apparently, the problem is not that easy after all. What your teacher believes is the solution is obviously wrong. After all, she suggests that $$\frac12$$ is one of the solutions, but $$(\tfrac12)^2-(k+1)\tfrac12+k+1\ne 0$$

• I was going crazy trying to get what she got. Thank you 🙏🏻 – Sophia Serrato Apr 19 '19 at 10:03

Hint $$:$$ $$(-k-1)^2 = (k+1)^2 =k^2 +2k+1.$$

• My mistake! Thank you! – Sophia Serrato Apr 19 '19 at 10:03

Your choice of parameters are correct so you should get that $$\sf{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{(k+1)\pm\sqrt\Delta}{2}}$$ where $$\sf{\Delta=(-(k+1))^2-4(1)(k+1)=k^2+2k+1-4k-4=k^2-2k-3\ne k^2}$$ so your teacher is wrong nonetheless.

• @farruhota Thanks, I misread the minus signs in the OP's attempt. – TheSimpliFire Apr 19 '19 at 10:03
• Thank you! I appreciate your help! 🙏🏻 – Sophia Serrato Apr 19 '19 at 10:06

It is often a good idea to check the results by numerical examples.

If we set $$k=1$$, the equation becomes $$x^2-x+1=0$$, and your teacher's answer would be

$$x=\frac12\ !?$$

Now,

$$\frac{(k+1)\pm\sqrt{(k+1)^2-4(k+1)}}2=\frac{(k+1)\pm\sqrt{(k+1)(k-3)}}2$$

yields

$$\frac{1\pm i\sqrt3}2,$$ which works.

A simpler check is possible with $$k=3$$, giving $$x=2$$ indeed a solution of $$x^2-4x+4=0$$.

A check is not a proof, but it can increase your trust in a solution.

• Thank you! I will do that next time before writing the answer my teacher has given me. – Sophia Serrato Apr 19 '19 at 10:26
• @SophiaSerrato: and after providing your own ! – Yves Daoust Apr 19 '19 at 10:28