I completed my high-school maths test and got $96$% which I was very proud of (not meaning to boast). But there was one question which I got $2$ marks on instead of $4$ and didn't know why when I looked back at my results.
$\tag{4 marks}$ Question 8:
Derive $y^2 - z^2$ from $(y + z)^2$ and then convert it into the quadratic form: $$ax^2 + bx + c = 0 : (a, b, c, x) \in \mathbb{R}, \ a\neq 0, \ x = y + z$$ How must you then show that: $$(y + z)^2 = \frac{-b + \sqrt{\Delta}}{2a}$$ and derive $y^2 - z^2$ from the $RHS$? What is the value of $\Delta$?
My Attempt:
I know that $\Delta = \text{discriminant}, b^2 - 4ac$. I also know that $RHS = \text{Right Hand Side}$. I also know that the numerator of the quadratic formula does not include $\pm$ because $\forall n\in \mathbb{R}, \ n^2 > 0$ meaning that there cannot exist $-\sqrt{\Delta}$. But I also know how to expand $(y + z)^2$ $$\begin{align} (y + z)^2 &= y^2 + z^2 + 2yz \\ \implies y^2 - z^2 &= (y + z)^2 - 2yz - 2z^2 \\ &= (y + z)^2 - (2yz + 2z^2) \\ &= (y + z)^2 - 2(yz + z^2) \\ &= (y + z)^2 - 2z(y + z) \\ \implies \frac{y^2 - z^2}{2z} &= \frac{(y + z)^2 - 2z(y + z)}{2z} \\ &= \frac{(y + z)^2}{2z} - \frac{\require{cancel}\cancel{2z}(y + z)}{\cancel{2z}} \\ &= \frac{(y + z)^2}{2z} - (y + z) \\ \implies 0 &= \frac{(y + z)^2}{2z} - (y + z) - \frac{y^2 - z^2}{2z} \\ &= \frac{1}{2z}(y + z)^2 + (-1)(y + z) + \bigg(-\frac{y^2 - z^2}{2z}\bigg) \rightarrow 0 = ax^2 + bx + c \\ \therefore y + z &= \frac{1 + \sqrt{1 - 4\times \frac{1}{2z}\times \big(-\frac{y^2 - z^2}{2z}\big)}}{2\times \frac{1}{2z}} \\ &= \frac{1 + \sqrt{1 + \cancel{2z}\big(\frac{y^2 - z^2}{\cancel{2z}}\big)}}{\frac{\cancel{2}}{\cancel{2}z}} \\ &= \frac{1 + \sqrt{1 + y^2 - z^2}}{z} \\ \end{align}$$ However, it is here that I stumble into a problem: $$\because y + z = z + y \implies \frac{1 + \sqrt{1 + y^2 - z^2}}{z} = \frac{1 + \sqrt{1 + z^2 - y^2}}{y} \implies y = z$$ and thus it is incorrect and I cannot continue from here.
Correction:
I asked my maths teacher where I went wrong, and he said that in the equation: $$y^2 - z^2 = (y + z)^2 - 2z(y + z)$$ I made the mistake of dividing both sides by $2z$, when from this I should have subtracted $y^2 - z^2$ to get $0$ already so that: $$\begin{align} 0 &= (y + z)^2 - 2z(y + z) - (y^2 - z^2) \\ \therefore y + z &= \frac{2z + \sqrt{(-2z)^2 + 4(y^2 - z^2)}}{2} \\ &= \frac{2z + \sqrt{4z^2 + 4(y^2 - z^2)}}{2} \\ &= \frac{2z + \sqrt{4(z^2 + y^2 - z^2)}}{2} \\ &= \frac{2z + \sqrt{4y^2}}{2} \\ &= \frac{2z + \sqrt{4}\sqrt{y^2}}{2} \\ &= \frac{2z + 2y}{2} \\ &= \frac{\cancel{2}(z + y)}{\cancel{2}} \\ &= y + z \\ \end{align}$$ and thus I can continue from here, deriving $y^2 - z^2$ from the quadratic formula. But what I do not understand is: $$\frac{y^2 - z^2}{2z} = \frac{(y + z)^2 - 2z(y + z)}{2z} \iff y^2 - z^2 = (y + z)^2 - 2z(y + z)$$ $$\therefore 0 = \frac{(y + z)^2}{2z} - (y + z) - \frac{y^2 - z^2}{2z} \iff 0 = (y + z)^2 - 2z(y + z) - (y^2 - z^2)$$ So why am I not getting the desired result? Could you please explain why? I would mostly appreciate it if you did not use different mathematical techniques in answering my question as opposed to quadratic techniques since this test was chiefly concerned on understanding quadratic equations (including completing the square and graphing). Other mathematical techniques could also be foreign to my understanding of mathematics so far; compared to most users here, I am not highly skilled in mathematics. If you are willing to not (just) include quadratic equations as a way of answering my question, please be very clear and show me step by step.
Thank you in advance.
First of all, when you divide by $2z$, you automatically exclude the case when $z=0$ and this case must be done separately.
But to the main question. The problem is that $$ \frac{1+\sqrt{1-4\times\frac{1}{2z}\times\left(-\frac{y^2-z^2}{2z}\right)}}{2\times \frac{1}{2z}} \not= \frac{1+\sqrt{1+2z\left(\frac{y^2-z^2}{2z}\right)}}{\frac{2}{2z}}, $$ but $$ \frac{1+\sqrt{1-4\times\frac{1}{2z}\times\left(-\frac{y^2-z^2}{2z}\right)}}{2\times \frac{1}{2z}} = \frac{1+\sqrt{1+\frac{y^2-z^2}{z^2}}}{\frac{2}{2z}}, $$ that is of course equal to $z+y$ for $y\geq 0$. But again, in this scenario, you exclude the case $z=0$.