The quadratic formula
The quadratic formula gives the exact roots of a quadratic equation in zero form. Calculate the discriminant first, then substitute the signed coefficients and simplify both branches
- Write zero form Write the equation in zero form before reading the three coefficients \(\displaystyle 2x^{2}-5x+2=0\)
- Identify signed coefficients Keep the sign attached to b and c when reading the equation \(\displaystyle a=2,\quad b=-5,\quad c=2\)
- Calculate the discriminant Use the signed coefficients to calculate the discriminant, then simplify exactly \(\displaystyle \Delta=b^{2}-4ac=(-5)^{2}-4(2)(2)=9\)
- Use both branches Keep the whole numerator over 2a and use both signs when the roots are real \(\displaystyle x=\frac{-b\pm\sqrt{\Delta}}{2a}\)
Worked example
Solve the quadratic equation exactly
- \(\displaystyle a=1,\quad b=-7,\quad c=10\)Read the signed coefficients from the equation in zero form
- \(\displaystyle \Delta=(-7)^{2}-4(1)(10)=49-40=9\)Square the signed value of b, then subtract 4ac
- \(\displaystyle x=\frac{-(-7)\pm\sqrt{9}}{2(1)}=\frac{7\pm3}{2}\)Substitute the coefficients with the whole numerator over 2a
- \(\displaystyle x=2\quad\text{or}\quad x=5\)The two branches give (7 − 3) ÷ 2 = 2 and (7 + 3) ÷ 2 = 5.
Common mistakes
Difficulty levels
Questions about this topic
Why must the equation be in zero form?+
The coefficients are defined by the equation in zero form, so all terms must be on one side before they are read
What does the discriminant tell you?+
A negative discriminant gives no real roots, zero gives one repeated real root, and a positive discriminant gives two real roots
Why does the formula contain plus-minus?+
The principal square root of a positive discriminant is non-negative. Both its positive and negative values solve the corresponding squared equation, so the plus-minus sign retains both branches
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 11: Binomial distribution