NDA · Chapter

Quadratic Equations

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for

ax^2 + bx + c = 0

,

a \neq 0

Start

The discriminant and the nature of roots

D = b^2 - 4ac

:

D > 0

two real roots,

D = 0

equal roots,

D < 0

no real roots; rational roots need

D

a perfect squareLocked

Sum and product of the roots $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$ , read off without solvingLocked
Expressions in the rootsRewrite $\alpha^2 + \beta^2$ , $\alpha^3 + \beta^3$ ,
Forming an equation from its roots $x^2 - (\text{sum})x + \text{product} = 0$ ; for new roots like $\frac{1}{\alpha}$
Relations between the rootsA condition such as "one root is twice the other" or "the roots are reciprocal" becomes an equation in the coefficientsLocked
Common rootsA shared root satisfies both equations; both roots shared means the coefficients are in proportionLocked

Sign and extreme value of a quadratic $ax^2 + bx + c > 0$ for every $x$ exactly when $a > 0$ and $D < 0$ ; its least (or greatest) value is $-\frac{D}{4a}$ Locked
Roots of given signsBoth roots positive needs $D \ge 0$ , sum $> 0$ , product $> 0$ ; roots of opposite signs need only product $< 0$
Equations that reduce to quadraticsSubstitute $t = 2^x$ , $t = \log x$ , $t = x + \frac{1}{x}$

(\alpha - \beta)^2

and

\frac{1}{\alpha} + \frac{1}{\beta}

in terms of

\alpha + \beta

and

\alpha\beta

Locked

or

\alpha^2

,

find their sum and product first

Locked

or

t = |x|

,

solve the quadratic in

t

,

then keep only the values

t

can take

Locked