Quadratic equations

A quadratic equation is an equation of the form

ax2+bx+c=0\boxed{ax^2+bx+c=0}

where a,ba, b and cc are called coefficients, and are real numbers. In fact, we also require a≠0a\neq 0, because for a=0a=0 we get a linear equation: bx+c=0bx+c=0. There is nothing wrong with this, we just do not call it a quadratic equation any more. If bb or cc are zero, we still get a quadratic equation (ax2+bx=0ax^2+bx=0 or ax2+c=0ax^2+c=0).

Determining the coefficients of a quadratic equation will prove to be important later, so let's discuss it a bit more carefully. We note three points:

  1. As shown above, we use the convention that aa multiplies the x2x^2-term, bb the xx-term, and cc is the constant term. For example, for the quadratic equation

    4x+3+2x2=04x+3+2x^2=0

    we have a=2,b=4a=2, b=4, and c=3c=3. And for the quadratic equation

    −2x2−4x−3=0-2x^2-4x-3=0

    the coefficients are a=−2,b=−4a=-2, b=-4 and c=−3c=-3.

  2. Important! For determining the coefficients a,ba, b and cc, we first have to collect all x2x^2-term, all xx-terms and all constant, and bring them on one side of the equation, so that on the other side of the equation is 00. For example, to find the coefficients of the quadratic equation

    2x2−5x+3−x−4=5−x2+10x−2.5+12x^2-5x+3-x-4=5-x^2+10x-2.5+1

    we first write

    3x2−16x−4.5=03x^2-16x-4.5=0

    and thus we see that a=3,b=−16a=3, b=-16, and c=−4.5c=-4.5. This last form, ax2+bx+c=0ax^2+bx+c=0, is called the standard form of the quadratic equation. Later we will learn about another useful form, the so called vertex form.

  3. A coefficient of a=1a=1 or b=1b=1 or b=0b=0 or c=0c=0 is normally omitted. E.g

    x2+x=0→a=1,b=1,c=0x^2+x=0 \rightarrow a=1, b=1, c=0

    or

    −x2=0→a=−1,b=0,c=0-x^2=0 \rightarrow a=-1, b=0, c=0
Exercise 1

Determine if the following equations are quadratic equations. If so, bring them into standard form and determine the coefficients aa, bb, and cc.

  1. 2x2−3x−3=02x^2-3x-3=0

  2. x+2=0x+2=0

  3. 3x2+x+3−5x+2=03x^2+x+3-5x+2=0

  4. (x−2)2=−4(x-2)^2=-4

  5. x(x+2)=0x(x+2)=0

  6. (x+3)(x−1)=2x2−4x+10(x+3)(x-1)=2x^2-4x+10

  7. (2x+1)(3x−4)=4(2x+1)(3x-4)=4

  8. x4(1−2x2)=3x^4(1-\frac{2}{x^2})=3

  9. x−x2+1=0\sqrt{x}-x^2+1=0

  10. 3x(1−x)=1−x3x(1-x)=1-x

  11. (3x−1)2=4(3x-1)^2=4

  12. −x2+3=0-x^2+3=0

  13. −x2=x-x^2=x

Solution
  1. yes, 2x2−3x−3=0→a=2,b=−3,c=−32x^2-3x-3=0 \rightarrow a=2, b=-3, c=-3
  2. no, linear.
  3. yes, 3x2−4x+5=0→a=3,b=−4,c=53x^2-4x+5=0 \rightarrow a=3, b=-4, c=5
  4. yes, x2−4x+8=0→a=1,b=−4,c=8x^2-4x+8=0 \rightarrow a=1, b=-4, c=8
  5. yes, x2+2x=0→a=1,b=2,c=0x^2+2x=0 \rightarrow a=1, b=2, c=0
  6. yes, x2+2x−3=2x2−4x+10→−x2+6x−13=0→a=−1,b=6,c=−13x^2+2x-3=2x^2-4x+10 \rightarrow -x^2+6x-13=0 \rightarrow a=-1, b=6, c=-13
  7. yes, 6x2−5x−8=0→a=6,b=−5,c=−86x^2 -5x-8=0 \rightarrow a=6, b=-5, c=-8
  8. no
  9. no
  10. yes, −3x2+4x−1=0→a=−3,b=4,c=−1-3x^2+4x -1= 0 \rightarrow a=-3, b=4, c=-1
  11. yes, 9x2−6x−3=0→a=9,b=−6,c=−39x^2-6x-3=0 \rightarrow a=9, b=-6, c=-3
  12. yes, −x2+3=0→a=−1,b=0,c=3-x^2+3=0 \rightarrow a=-1, b=0, c=3
  13. yes, −x2−x=0→a=−1,b=−1,c=0-x^2-x=0 \rightarrow a=-1, b=-1, c=0

So, how do we solve quadratic equations? In general, this is tough. For example, consider this equation:

x2−2x−1=0x^2-2x-1=0

What we could try is to get the x2x^2 on the left side, and everything else on the right side, and then take the root:

x2−2x−1=0∣−2x,+1x2=2x+1∣xx=±2x+1\begin{array}{lll} x^2-2x-1&=&0 &\quad | -2x, +1\\ x^2&=& 2x+1 &\quad | \sqrt{\phantom{x}}\\ x&=&\pm \sqrt{2x+1} \end{array}

But this doesn't really help for finding a value for xx, because xx still appears on both sides of the equation. So, we need a new method for solving quadratic equations. We will learn about two such methods. But before we discuss those in the next sections, let us finish this section by repeating ways to solve specific types of quadratic equations (see section 16):

Exercise 2

Solve the following quadratic equations. Also, what are its coefficients a,ba, b and cc?

  1. 4x2−1=04x^2-1=0

  2. (x−2)(2x+1)=0(x-2)(2x+1)=0

Solution

  1. The coefficients are a=4,b=0a=4, b=0 and c=−1c=-1. The solution is:

    4x2−1=0∣+1,:4x2=14∣xx1,2=±12\begin{array}{rlll} 4x^2-1&=&0&\quad | +1, :4\\ x^2&=& \frac{1}{4}& \quad| \sqrt{\phantom{x}}\\ x_{1,2}&=&\pm \frac{1}{2} \end{array}

  2. To find the coefficients, we expand:

    (x−2)(2x+1)=2x2+x−4x−2=2x2−3x−2(x-2)(2x+1)=2x^2+x-4x-2=2x^2-3x-2

    and thus we see that a=2,b=−3a=2, b=-3 and c=−2c=-2.

    The equation

    (x−2)(2x+1)=0(x-2)(2x+1)=0

    can be solve by observing that the first factor has to 00

    x−2=0x-2=0

    or the second factor has to be 00

    2x+1=02x+1=0

    Thus it follows x1=2x_1=2 and x2=−12x_2=-\frac{1}{2}.