# Quadratic Equation Formula and the Discriminant

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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.

1. Start from the equation ax^2+bx+c=0.
2. Divide both sides by a: x^2+b/ax+c/a=0.
3. Move constant term to the right: x^2+b/ax=-c/a.
4. Add (b/2a)^2=b^2/(4a^2) to both sides of the equation: x^2+b/ax+b^2/(4a^2)=-c/a+b^2/(4a^2).
5. Rewrite left hand side: (x+b/(2a))^2=-c/a+b^2/(4a^2).
6. Simplify right hand side: -c/a+b^2/(4a^2)=(-c*color(red)(4a))/(a*color(red)(4a))+b^2/(4a^2)=(-4ac)/(4a^2)+b^2/(4a^2)=(b^2-4ac)/(4a^2).
7. Write the final equation: (x+b/(2a))^2=(b^2-4ac)/(4a^2).
8. Solve the equation: x+b/(2a)=sqrt((b^2-4ac)/(4a^2)) or x+b/(2a)=-sqrt((b^2-4ac)/(4a^2)).
9. Above equations have roots x_1=(-b+sqrt(b^2-4ac))/(2a) and x_2=(-b-sqrt(b^2-4ac))/(2a).
10. We can write it even more compactly: x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a).

Expression b^2-4ac is called the discriminant of the quadratic equation.

Since it is under square root, then, depending on its value, quadratic equation will have 0, 1 or 2 roots.

Indeed, if b^2-4ac>0, then sqrt(b^2-4ac) exists and we will get two solutions.

If b^2-4ac=0, then +-sqrt(b^2-4ac)=+-sqrt(0)=0 and we will get one solution.

If b^2-4ac<0, then there are no real roots, because square root of negative number doesn't exist.

Quadratic Equation Formula: color(brown)(ax^2+bx+c=0) has following roots: color(magenta)(x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)).

color(green)(D=b^2-4ac) is called the Discriminant of the quadratic equation.

• If b^2-4ac>0, equation has 2 real roots.
• If b^2-4ac=0, equation has 1 real root.
• If b^2-4ac<0, equation has no real roots.

Let's go through all three cases.

Example 1. Solve the equation using quadratic equation formula: 5x^2-9x-2=0.

Here a=5, b=-9, c=-2.

Calculate discriminant: D=(-9)^2-4*5*(-2)=81-(-40)=121.

Since discriminant is positive, then we expect 2 real roots.

x_1=(-(-9)+sqrt(121))/(2*5)=(9+11)/10=20/10=2.

x_2=(-(-9)-sqrt(121))/(2*5)=(9-11)/10=-2/10=-1/5.

Answer: roots are 2 and -1/5.

However, we will not always get rational roots. Roots can also be irrational.

Example 2. Solve the following equation: y^2-6y+2=0.

Here a=1, b=6, c=2.

Calculate discriminant: D=6^2-4*1*2=36-8=28.

Since discriminant is positive, then we expect 2 real roots.

x_1=(-6+sqrt(28))/(2*1)=(-6+sqrt(4*7))/2=(-6+sqrt(4)*sqrt(7))/2=(-6+2sqrt(7))/2=-6/2+(2sqrt(7))/2=-3+sqrt(7).

x_2=(-6-sqrt(28))/(2*1)=(-6-sqrt(4*7))/2=(-6-sqrt(4)*sqrt(7))/2=(-6-2sqrt(7))/2=-3-sqrt(7).

Answer: roots are -3+sqrt(7) and -3-sqrt(7).

Next case is when there is only one root.

Example 3. Solve 9x^2=-6x-1.

First of all, write equation in standard form: 9x^2+6x+1=0.

Here a=9, b=6, c=1.

Calculate discriminant: D=6^2-4*9*1=0.

Since discriminant is 0, then we expect 1 real root.

x=(-6+sqrt(0))/(2*9)=-6/18=-1/3.

Answer: root is -1/3.

Finally, consider a case, when there are no real roots.

Example 4. Solve the equation, using quadratic equation formula: 9c^2+6c=-2.

First of all, write equation in standard form: 9c^2+6c+2=0.

Here a=9, b=6, c=2.

Calculate discriminant: D=6^2-4*9*2=-36.

Since discriminant is negative, then there are no roots.

Quadratic equation can be used to solve incomplete quadratic equations (although, there are faster methods).

Example 5. Solve incomplete quadratic equation -3x^2+5=0.

Here a=-3, b=0, c=5.

Calculate discriminant: D=0^2-4*(-3)*5=60.

x_1=(-0+sqrt(60))/(2*(-3))=-sqrt(60)/6=-sqrt(60)/sqrt(36)=-sqrt(60/36)=-sqrt(5/3).

x_2=(-0-sqrt(60))/(2*(-3))=sqrt(5/3).

Answer: roots are -sqrt(5/3) and sqrt(5/3).

Now, it is time to exercise.

Exercise 1. Solve the following incomplete equation, using qudratic formula: 2x^2-5/3x=0.

Answer: 0 and 5/6.

Exercise 2. Solve equation 24x^2=-7x+5.

Answer: -5/8 and 1/3.

Exercise 3. Find roots of the equation 2c^2-3c=-21.

Exercise 4. Solve 1/4x^2-5x+25=0.
Answer: 10.
Exercise 5. Solve the following euqation, using qudratic formula: 1/3y^2+5y-7=0.
Answer: (-15-sqrt(309))/2 and (-15+sqrt(309))/2.