Quadratic equation in one variable is the equation with standard form a x 2 + b x + c = 0 \color{purple}{{{a}{{x}}^{{2}}+{b}{x}+{c}={0}}} a x 2 + b x + c = 0
a a a b b b c c c x x x a a a
Quadratic equation a x 2 + b x + c = 0 {a}{{x}}^{{2}}+{b}{x}+{c}={0} a x 2 + b x + c = 0 b {b} b c {c} c
Such equations can be easily solved without advanced methods.
Example 1. Solve x 2 − 81 = 0 {{x}}^{{2}}-{81}={0} x 2 − 81 = 0
Completing the square is a method for solving quadratic equation.
Before we into the method itself, let's start from a simple example.
Example 1. Solve equation ( x − 3 ) 2 = 16 {{\left({x}-{3}\right)}}^{{2}}={16} ( x − 3 ) 2 = 16
Quadratic Equation Formula can be derived from the steps for completing the square (actually, this formula is a general case).
Let's see how to do it.
Start from the equation a x 2 + b x + c = 0 {a}{{x}}^{{2}}+{b}{x}+{c}={0} a x 2 + b x + c = 0 Divide both sides by a {a} a x 2 + b a x + c a = 0 {{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}={0} x 2 + a b x + a c = 0 Move constant term to the right: x 2 + b a x = − c a {{x}}^{{2}}+\frac{{b}}{{a}}{x}=-\frac{{c}}{{a}} x 2 + a b x = − a c Add ( b 2 a ) 2 = b 2 4 a 2 {{\left(\frac{{b}}{{2}}{a}\right)}}^{{2}}=\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}} ( 2 b a ) 2 = 4 a 2 b 2 x 2 + b a x + b 2 4 a 2 = − c a + b 2 4 a 2 {{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}} x 2 + a b x + 4 a 2 b 2 = − a c + 4 a 2 b 2 Rewrite left hand side: ( x + b 2 a ) 2 = − c a + b 2 4 a 2 {{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}} ( x + 2 a b ) 2 = − a c + 4 a 2 b 2 Simplify right hand side: − c a + b 2 4 a 2 = − c ⋅ 4 a a ⋅ 4 a + b 2 4 a 2 = − 4 a c 4 a 2 + b 2 4 a 2 = b 2 − 4 a c 4 a 2 -\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{c}\cdot{\color{red}{{{4}{a}}}}}}{{{a}\cdot{\color{red}{{{4}{a}}}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}} − a c + 4 a 2 b 2 = a ⋅ 4 a − c ⋅ 4 a + 4 a 2 b 2 = 4 a 2 − 4 a c + 4 a 2 b 2 = 4 a 2 b 2 − 4 a c Write the final equation: ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 {{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}} ( x + 2 a b ) 2 = 4 a 2 b 2 − 4 a c Solve the equation: x + b 2 a = b 2 − 4 a c 4 a 2 {x}+\frac{{b}}{{{2}{a}}}=\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}} x + 2 a b = 4 a 2 b 2 − 4 a c x + b 2 a = − b 2 − 4 a c 4 a 2 {x}+\frac{{b}}{{{2}{a}}}=-\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}} x + 2 a b = − 4 a 2 b 2 − 4 a c Above equations have roots x 1 = − b + b 2 − 4 a c 2 a {x}_{{1}}=\frac{{-{b}+\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}} x 1 = 2 a − b + b 2 − 4 a c x 2 = − b − b 2 − 4 a c 2 a {x}_{{2}}=\frac{{-{b}-\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}} x 2 = 2 a − b − b 2 − 4 a c We can write it even more compactly: x 1 , 2 = − b ± b 2 − 4 a c 2 a {x}_{{{1},{2}}}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}} x 1 , 2 = 2 a − b ± b 2 − 4 a c Expression b 2 − 4 a c {{b}}^{{2}}-{4}{a}{c} b 2 − 4 a c
Viet Theorem. If quadratic equation a x 2 + b x + c = 0 {a}{{x}}^{{2}}+{b}{x}+{c}={0} a x 2 + b x + c = 0 x 2 + b a + c a = 0 {{x}}^{{2}}+\frac{{b}}{{a}}+\frac{{c}}{{a}}={0} x 2 + a b + a c = 0 p {p} p q {q} q p + q = − b a {\color{green}{{{p}+{q}=-\frac{{b}}{{a}}}}} p + q = − a b {\color{ma\genta}{{{p}{q}=\frac{{c}}{{a}}}}}
