Lesson Solving quadratic equations without quadratic formula

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Solving quadratic equations without quadratic formula


Lessons PROOF of quadratic formula... and Introduction into quadratic equations of this module explain what is the quadratic formula and how to use it to solve quadratic equations.

Actually, the quadratic formula is like "heavy tank" and it works always.

But in some simple cases you can solve a quadratic equation easier and faster without using this formula, and this lesson explains how to do it.

First form of the quadratic equation


Let us consider quadratic equation x%5E2+=+4. The solution is obvious: x=2 and x=-2 (two roots).
Same in the case x%5E2+=+9. The solution is x=3 and x=-3.

What about x%5E2+=+25?
x%5E2+=+49?
x%5E2+=+36?
x%5E2+=+81?
x%5E2+=+64?
x%5E2+=+6.25?

x%5E2+=+5?
The solution is x=sqrt%285%29 and x=-sqrt%285%29.

x%5E2+=+5%2Bsqrt%283%29?
The solution is x=sqrt%285%2Bsqrt%283%29%29 and x=-sqrt%285%2Bsqrt%283%29%29.

And so on ...

Now please solve yourself similar equations:
x%5E2=121.
x%5E2=15.
x%5E2=9%2Bsqrt%287%29.

Second form of the quadratic equation


Now let us consider quadratic equation %283x%2B2%29%5E2=25.
This one also can be solved simply.
Note first that taking square root from both sides we have 3x+2=5 or 3x+2=-5.
From the first of these two equations we have 3x=5-2=3 and x=1.
From the second equation 3x=-5-2=-7 and x=-7/3.
The solutions: x=1 and x=-7/3.

Now please solve yourself similar equations:
%287x-3%29%5E2=4
%2812x%2B5%29%5E2=11
%283x%2B4%29%5E2=7%2Bsqrt%285%29

Using Viete's formulas


Let us consider the quadratic equation

a%2Ax%5E2+%2Bb%2Ax+%2B+c+=+0.                         (1)

If x%5B1%5D and x%5B2%5D are roots of the quadratic equation (1), then

x%5B1%5D+%2B+x%5B2%5D+=+-b%2Fa and x%5B1%5D%2Ax%5B2%5D+=+c%2Fa.   (2)

Formulas (2) are Vieta's formulas for quadratic equation (1) (see Viete's formulas in Wikipedia or the proof at the end of this section).

Viete's formulas are usually applied to the quadratic equation with the coefficient a at x%5E2 equal to 1. In this case these formulas become particularly simple:

x%5B1%5D%2Bx%5B2%5D+=+-b and x%5B1%5D%2Ax%5B2%5D+=+c.   (3)

Now, if you are given the quadratic equation

x%5E2%2Bpx%2Bq=0                             (4)

with the coefficient at the x%5E2 term equal to 1, try to guess right two numbers x%5B1%5D and x%5B2%5D that match to conditions

x%5B1%5D%2Ax%5B2%5D+=+q and x%5B1%5D%2Bx%5B2%5D=-p.   (5)

If you succeed, these numbers, x%5B1%5D and x%5B2%5D, are solutions of the quadratic equation (4).

Example 1
Solve quadratic equation x%5E2-4x%2B3=0 using Viete's formulas.
Note that 3*1=3 and 3+1=4, so numbers 3 and 1 are solutions, according to (5).
Surely, because we only made guess right, we should check these solutions directly and independently. Please do it yourself.

Example 2
Solve quadratic equation x%5E2%2B2x-3=0 using Viete's formulas.
Note that (-3)*1=-3 and (-3)+1=-2, so numbers -3 and 1 are solutions, according to (5).
Again, because we only made guess right, we should check these solutions directly and independently. Please do it yourself.

Example 3
Solve quadratic equation x%5E2-5x%2B6=0 using Viete's formulas.
Note that 3*2=6 and 3+2=5, so numbers 3 and 2 are solutions, according to (5).
You should check these solutions directly and independently. Please do it yourself.

More examples
Solve quadratic equations using Viete's formulas.
x%5E2-7x-8=0.
x%5E2%2B3x%2B2=0.
x%5E2%2B5x%2B6=0.
x%5E2%2B10x-24=0.
x%5E2%2B4x%2B3=0.
x%5E2%2B6x-7=0.
x%5E2-8x%2B15=0.
x%5E2-6x%2B8=0.

Don't forget check yours solutions directly and independently.

The proof of Viete's formulas


If x%5B1%5D and x%5B2%5D are roots of the quadratic equation (1), then
a%2Ax%5E2%2Bb%2Ax%2Bc+=+a%2A%28x-x%5B1%5D%29%2A%28x-x%5B2%5D%29. Performing multiplication of polynomials in the right side
and comparing coefficients separately at x and at constant terms at the left and the right side, we obtain formulas (2).


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