10th Problem 2016

Algebra Level 1

What is the quadratic formula for the equation a x 2 + b x + c = 0 ax^2 + bx + c = 0 ?

Check out the set: 2016 Problems

x = − b ± b 2 − 4 a c a x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2\quad }-\quad 4ac } }{ a } x = − b ± b − 4 a c 2 a x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ \quad }-\quad 4ac } }{ 2a } x = b ± b 2 − 4 a c 2 a x\quad =\quad \frac { b\quad \pm \quad \sqrt { { b }^{ 2\quad }-\quad 4ac } }{ 2a } x = − b ± b 2 − 4 a c 2 a x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2\quad }-\quad 4ac } }{ 2a }

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ananth Jayadev
Jan 1, 2016

Proof of the Quadratic Formula:

The original equation is a x 2 + b x + c = 0 ax^2 + bx + c = 0 .

Subtract the c c on both sides so a x 2 + b x = − c ax^2 + bx = -c .

Divide the a a by all terms so x 2 + ( b a ) x = − c a x^2 + (\frac {b}{a})x = -\frac {c}{a} .

Take half the x x term and square it so b 2 a → b 2 4 a 2 \frac {b}{2a} \rightarrow \frac {b^2}{4a^2} .

Add the squared term on both sides so 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}{4a^2} = -\frac {c}{a} + \frac {b^2}{4a^2} .

Simplify the RHS so x 2 + ( b a ) x + b 2 4 a 2 = − 4 a c 4 a 2 + b 2 4 a 2 x^2 + (\frac {b}{a})x + \frac {b^2}{4a^2} = -\frac {4ac}{4a^2} + \frac {b^2}{4a^2} .

Convert the LHS to square form so ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 (x+\frac {b}{2a})^2 = \frac {b^2 - 4ac}{4a^2} .

Square root both sides so x + b 2 a = ± b 2 − 4 a c 4 a 2 = ± b 2 − 4 a c 2 a x+\frac { b }{ 2a } =\pm \sqrt { \frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } } =\quad \frac { \pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } .

Solve for x x so x = − b 2 a ± b 2 − 4 a c 2 a = − b ± b 2 − 4 a c 2 a x = -\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } = \frac{-b \pm \sqrt{b^2-4ac}}{2a} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Derived using Completing the Square.. Amazing :D

Angela Fajardo - 5 years, 5 months ago

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