A quadratic equation is an equation of the form $$ax^2+bx+c=0,\quad a\neq 0,$$ where $a$, $b$, $c$ are real numbers: $a$ is the leading coefficient, $b$ the linear coefficient, $c$ the constant term. The values of $x$ that satisfy the equation are called its roots, and the set of all roots is the solution set. Before solving, always bring the equation to standard form. Special cases: if $b=0$, then $ax^2+c=0\Rightarrow x^2=-\frac{c}{a}$ (e.g. $x^2=4\Rightarrow x=\pm 2$, solution set $\{-2,\,2\}$); if $c=0$, then $x(ax+b)=0\Rightarrow x=0$ or $x=-\frac{b}{a}$.
For $ax^2+bx+c=0$ ($a\neq 0$) the discriminant is $$\Delta=b^2-4ac.$$ The roots are given by the quadratic formula: $$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}.$$ Number of real roots: if $\Delta>0$ — two distinct real roots; if $\Delta=0$ — one repeated (double) root $x_1=x_2=-\frac{b}{2a}$; if $\Delta<0$ — no real roots (solution set over $\mathbb{R}$ is $\varnothing$). Example: for $x^2+x-1=0$ we get $\Delta=1^2-4\cdot 1\cdot(-1)=5>0$, so $x_{1,2}=\frac{-1\pm\sqrt{5}}{2}$ and the solution set is $\left\{\frac{-1-\sqrt{5}}{2},\,\frac{-1+\sqrt{5}}{2}\right\}$.
If $x_1$ and $x_2$ are the roots of $ax^2+bx+c=0$ ($a\neq 0$), then Vieta's theorem states: $$x_1+x_2=-\frac{b}{a},\qquad x_1\cdot x_2=\frac{c}{a}.$$ This lets you work with the roots without finding them explicitly — the key tool in YÖS problems. For the reduced equation $x^2+px+q=0$: $x_1+x_2=-p$, $x_1x_2=q$. Note: Vieta's formulas hold whenever the roots exist (for real roots require $\Delta\ge 0$). Signs of the roots: if $\frac{c}{a}>0$, the roots have the same sign; if $\frac{c}{a}<0$ — opposite signs.
Let $S=x_1+x_2$ and $P=x_1x_2$ (from Vieta). The most common expressions: $$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=S^2-2P,$$ $$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=\frac{S}{P},$$ $$(x_1-x_2)^2=S^2-4P,\qquad x_1^3+x_2^3=S^3-3PS.$$ Non-symmetric combinations are also reduced to $S$ and $P$: e.g. $$x_1+\frac{1}{x_2}=\frac{x_1x_2+1}{x_2}.$$ If the problem gives such a value for a specific root, use $x_2=\frac{P}{x_1}$ or substitute the root into the equation. If the coefficients contain the roots themselves (e.g. $x^2+(3x_2-5)x+2x_1=0$), write Vieta's system: $x_1+x_2=-(3x_2-5)$ and $x_1x_2=2x_1$, then solve it for $x_1$ and $x_2$.
A quadratic equation with roots $x_1$ and $x_2$ (and leading coefficient $1$) is $$x^2-(x_1+x_2)\,x+x_1x_2=0,\qquad\text{i.e.}\qquad x^2-Sx+P=0.$$ Equivalently, $a(x-x_1)(x-x_2)=0$ for any $a\neq 0$. Example: an equation with roots $2$ and $-5$: $S=-3$, $P=-10$, so $x^2+3x-10=0$. This formula also works in reverse: to build a new equation whose roots are, say, $x_1+1$ and $x_2+1$, compute the new sum $S'=S+2$ and the new product $P'=P+S+1$, then write $x^2-S'x+P'=0$.
Many YÖS problems fix the *type* of roots and ask for the parameter. Translate the condition into an inequality or equation for $\Delta$: two distinct real roots $\Leftrightarrow\Delta>0$; equal (double) roots $\Leftrightarrow\Delta=0$; no real roots $\Leftrightarrow\Delta<0$. Example (no real roots): $x^2-5mx+25=0$ has no real roots when $\Delta=25m^2-100<0\Rightarrow m^2<4\Rightarrow -2<m<2$; if $m$ must be an integer, the possible values are $-1,\,0,\,1$ and their sum is $0$. Example (equal roots): if $x_1=x_2=4$, then by Vieta $S=8$ and $P=16$, so the equation is $x^2-8x+16=0$; equivalently, require $\Delta=0$ together with $-\frac{b}{2a}=4$. Always check the extra condition $a\neq 0$ if the leading coefficient contains the parameter.
If a relation between the roots is given (e.g. $x_1=3x_2$, $x_1-x_2=d$, $\frac{x_1}{x_2}=k$), combine it with Vieta's formulas into a system. Scheme for $x_1=3x_2$ in $x^2+bx+c=0$ with a parameter $m$ inside $b$ or $c$: from $x_1+x_2=4x_2=-b$ find $x_2$; then the product $x_1x_2=3x_2^2=c$ gives an equation for the parameter $m$. Example: for $x^2-8x+m=0$ with $x_1=3x_2$: $4x_2=8\Rightarrow x_2=2$, $x_1=6$, so $m=x_1x_2=12$. If instead a specific value of one root is known, simply substitute it into the equation — a root always satisfies its equation — and find the second root from Vieta.
Equations that are 'quadratic in disguise' are solved by a substitution $t=g(x)$. Biquadratic: $ax^4+bx^2+c=0\Rightarrow t=x^2$ (require $t\ge 0$). Exponential: in $4^x-9\cdot 2^x+8=0$ note that $4^x=(2^x)^2$, so with $t=2^x$ ($t>0$): $t^2-9t+8=0\Rightarrow t=1$ or $t=8\Rightarrow 2^x=1$ or $2^x=8\Rightarrow x=0$ or $x=3$. The same works for $9^x-28\cdot 3^x+27=0$ with $t=3^x$. Repeated block: in $\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1=0$ set $t=x+\frac{1}{x}$: $(t-1)^2=0\Rightarrow t=1$, then solve $x+\frac{1}{x}=1\Rightarrow x^2-x+1=0$, where $\Delta=-3<0$ — no real roots. Algorithm: substitute, solve the quadratic in $t$, discard the $t$-values violating the natural constraint ($t>0$ for $t=a^x$; $t\ge 0$ for $t=x^2$ or $t=\sqrt{x}$; $|t|\ge 2$ for $t=x+\frac{1}{x}$), then return to $x$.
Key identity: $x^2=|x|^2$. Hence an equation like $x^2-|x|-12=0$ becomes quadratic in $t=|x|$ with $t\ge 0$: $$t^2-t-12=0\Rightarrow (t-4)(t+3)=0\Rightarrow t=4\ \text{or}\ t=-3.$$ Since $t=|x|\ge 0$, discard $t=-3$; then $|x|=4\Rightarrow x=\pm 4$. The solution set is $\{-4,\,4\}$ and the product of the roots is $-16$. Alternative method — case analysis: for $x\ge 0$ replace $|x|$ by $x$, for $x<0$ by $-x$, solve each case and keep only the roots consistent with the case condition. The substitution $t=|x|$ is faster when the equation contains only $x^2$ and $|x|$.
Type 1 — quadratic in $\sqrt{x}$: in $x+\sqrt{x}-6=0$ set $t=\sqrt{x}$, $t\ge 0$: $t^2+t-6=0\Rightarrow t=2$ or $t=-3$; discard $t=-3$, so $\sqrt{x}=2\Rightarrow x=4$. Type 2 — $\sqrt{f(x)}=g(x)$: the equation is equivalent to the system $$f(x)=g(x)^2,\qquad g(x)\ge 0$$ (then $f(x)\ge 0$ holds automatically). Squaring is not an equivalent transformation — it can create *extraneous roots*, so checking the roots is mandatory. Example: $\sqrt{2x-1}=x-2$. Domain: $2x-1\ge 0\Rightarrow x\ge\frac{1}{2}$; we also need $x-2\ge 0\Rightarrow x\ge 2$. Squaring: $2x-1=x^2-4x+4\Rightarrow x^2-6x+5=0\Rightarrow x=1$ or $x=5$. Check: $x=1$ gives the right-hand side $1-2=-1<0$ — extraneous, reject; $x=5$: $\sqrt{9}=3=5-2$ — valid. Answer: $x=5$.
Problem. $x_1$ and $x_2$ are the roots of $x^2+x-1=0$. Find $x_1^2+x_2^2$ and $\frac{1}{x_1}+\frac{1}{x_2}$. Solution. Step 1. Check that real roots exist: $\Delta=1+4=5>0$. Step 2. By Vieta: $S=x_1+x_2=-1$, $P=x_1x_2=-1$. Step 3. $$x_1^2+x_2^2=S^2-2P=(-1)^2-2\cdot(-1)=1+2=3.$$ Step 4. $$\frac{1}{x_1}+\frac{1}{x_2}=\frac{S}{P}=\frac{-1}{-1}=1.$$ Answer: $x_1^2+x_2^2=3$, $\frac{1}{x_1}+\frac{1}{x_2}=1$. Note that we never computed the irrational roots themselves.
Problem. The roots of $2x^2-x+p^2+3p=0$ satisfy $x_1\cdot x_2=5$, where $p\in\mathbb{R}^+$. Find $p$. Solution. Step 1. By Vieta for $ax^2+bx+c=0$: $x_1x_2=\frac{c}{a}=\frac{p^2+3p}{2}$. Step 2. Set it equal to $5$: $$\frac{p^2+3p}{2}=5\Rightarrow p^2+3p-10=0.$$ Step 3. Solve for $p$: $\Delta=9+40=49$, so $p=\frac{-3\pm 7}{2}$, i.e. $p=2$ or $p=-5$. Step 4. The condition $p\in\mathbb{R}^+$ keeps only $p=2$. Answer: $p=2$.
Problem. Solve $9^x-28\cdot 3^x+27=0$ and find the sum of its roots. Solution. Step 1. Since $9^x=(3^x)^2$, substitute $t=3^x$, $t>0$: $$t^2-28t+27=0.$$ Step 2. By Vieta (or factoring): $(t-1)(t-27)=0\Rightarrow t=1$ or $t=27$; both are positive, so both are accepted. Step 3. Return to $x$: $3^x=1\Rightarrow x=0$; $3^x=27=3^3\Rightarrow x=3$. Step 4. Solution set: $\{0,\,3\}$; the sum of the roots is $3$. Answer: $x\in\{0,\,3\}$, sum $=3$. The equation $4^x-9\cdot 2^x+8=0$ is solved identically with $t=2^x$: $t=1$ or $t=8$, so $x=0$ or $x=3$.
Problem. Solve $x+\sqrt{x}-6=0$. Solution. Step 1. Domain: $x\ge 0$. Step 2. Substitute $t=\sqrt{x}$, so $x=t^2$ and $t\ge 0$: $$t^2+t-6=0.$$ Step 3. Factor: $(t+3)(t-2)=0\Rightarrow t=-3$ or $t=2$. Since $t=\sqrt{x}\ge 0$, discard $t=-3$. Step 4. Back to $x$: $\sqrt{x}=2\Rightarrow x=4$. Step 5. Check in the original equation: $4+\sqrt{4}-6=4+2-6=0$ — correct. Answer: $x=4$, solution set $\{4\}$. A typical trap is to treat $x=9$ (coming from $t=-3$, $t^2=9$) as a root — verify: $9+3-6=6\neq 0$, so it is extraneous. This is exactly why the constraint on $t$ and the final check are essential.
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