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If $c > a > 0$ and $a − b + c = 0$, find the largest root of $ax^2 + bx + c = 0$.

Edit: The key says that the larger root is $-a-b+c$. Is there something wrong with the key?

You have: $$ax^2+bx+c=0$$
When you put $x=-1$, you get: $$a-b+c=0$$
Roots are: $$frac{-bpmsqrt{b^2-4ac}}{2a}$$
Substitute $b=a+c$ in it: $$frac{-(a+c)pmsqrt{(a+c)^2-4ac}}{2a}$$
$$=frac{-(a+c)pmsqrt{(a-c)^2}}{2a}$$
Solving, you get roots as $-frac ca$ and $-1$, out of which, $-1$ will be greater,

because: $$c>a>0 implies frac ca>1$$

Using that $$b=a+c$$ and the quadratic formula we get
$$x_{1,2}=-frac{a+c}{2a}pmfrac{sqrt{(a-c)^2}}{2a}$$ since $$a>c$$ we get
$$x_{1,2}=-frac{a+c}{2a}pmfrac{a-c}{2a}$$

Since $a-b+c=0$, $-1$ is a root of $ax^2+bx+c=0$.

It is well-known that the product of the roots of a quadratic equation $ax^2+bx+c=0$ is $dfrac ca$, so the other root is $-dfrac ca$.

Since $c>a>0$, the other root is less than $-1$, hence $-1$ is the largest root.

For a quadratic equation $ax^2 + bx +c = 0$ with $a neq 0$, the roots are $x = frac{-b pmsqrt{b^2 - 4ac}}{2a}.$

We are given $b = a+c$. Substituting $b$ for $a+c$ in the expression for the roots yields $$x = frac{-(a+c) pm sqrt{(a-c)^2}}{2a} = frac{-(a+c) pm |a-c|}{2a}.$$ We know $c > a iff a -c < 0$, which implies that $|a-c|= -(a-c) = c-a$. Hence, $$x = frac{-a-c pm (c-a)}{2a} iff x = -1 text{or} x = -frac{c}{a}. $$ So, which one is larger? We can find out by taking the difference of the roots, $-1 - (-frac{c}{a}) = -1 + frac{c}{a} = frac{c-a}{a} > 0$ since $c > a$ and $a > 0$.

Hence, $(-1) - (-frac{c}{a}) > 0 iff -1 > -frac{c}{a}.$ I.e. $-1$ is the larger root.