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Equations and inequalities: Solving quadratic equations

A quadratic equation is an equation where the power of the variable is at most 2. The following are examples of quadratic equations.

2 x 2 + 2 x = 1 2 - x 3 x + 1 = 2 x 4 3 x - 6 = 7 x 2 + 2 2 x 2 + 2 x = 1 2 - x 3 x + 1 = 2 x 4 3 x - 6 = 7 x 2 + 2
(1)

Quadratic equations differ from linear equations by the fact that a linear equation only has one solution, while a quadratic equation has at most two solutions. However, there are some special situations when a quadratic equation only has one solution.

We solve quadratic equations by factorisation, that is writing the quadratic as a product of two expressions in brackets. For example, we know that:

( x + 1 ) ( 2 x - 3 ) = 2 x 2 - x - 3 . ( x + 1 ) ( 2 x - 3 ) = 2 x 2 - x - 3 .
(2)

In order to solve:

2 x 2 - x - 3 = 0 2 x 2 - x - 3 = 0
(3)

we need to be able to write 2x2-x-32x2-x-3 as (x+1)(2x-3)(x+1)(2x-3), which we already know how to do. The reason for equating to zero and factoring is that if we attempt to solve it in a 'normal' way, we may miss one of the solutions. On the other hand, if we have the (non-linear) equation f(x)g(x)=0f(x)g(x)=0, for some functions ff and gg, we know that the solution is f(x)=0f(x)=0 OR g(x)=0g(x)=0, which allows us to find BOTH solutions (or know that there is only one solution if it turns out that f=gf=g).

Investigation : Factorising a Quadratic

Factorise the following quadratic expressions:

  1. x + x 2 x + x 2
  2. x 2 + 1 + 2 x x 2 + 1 + 2 x
  3. x 2 - 4 x + 5 x 2 - 4 x + 5
  4. 16 x 2 - 9 16 x 2 - 9
  5. 4 x 2 + 4 x + 1 4 x 2 + 4 x + 1

Being able to factorise a quadratic means that you are one step away from solving a quadratic equation. For example, x2-3x-2=0x2-3x-2=0 can be written as (x-1)(x-2)=0(x-1)(x-2)=0. This means that both x-1=0x-1=0 and x-2=0x-2=0, which gives x=1x=1 and x=2x=2 as the two solutions to the quadratic equation x2-3x-2=0x2-3x-2=0.

Method: Solving Quadratic Equations

  1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form ax2+bx+c=0ax2+bx+c=0 where aa, bb and cc have no common factors. For example, 2x2+4x+2=02x2+4x+2=0
     
    can be written as x2+2x+1=0x2+2x+1=0
     
    by dividing by 2.
  2. Write ax2+bx+cax2+bx+c in terms of its factors (rx+s)(ux+v)(rx+s)(ux+v). This means (rx+s)(ux+v)=0(rx+s)(ux+v)=0.
  3. Once writing the equation in the form (rx+s)(ux+v)=0(rx+s)(ux+v)=0, it then follows that the two solutions are x=-srx=-sr or x=-uvx=-uv.
  4. For each solution substitute the value into the original equation to check whether it is valid

Solutions of Quadratic Equations

There are two solutions to a quadratic equation, because any one of the values can solve the equation.

Figure 1
Khan academy video on equations - 3

Exercise 1: Solving Quadratic Equations

Solve for xx: 3x2+2x-1=03x2+2x-1=0

Solution
  1. Step 1. Find the factors of 3x2+2x-13x2+2x-1 :

    As we have seen the factors of 3x2+2x-13x2+2x-1 are (x+1)(x+1) and (3x-1)(3x-1).

  2. Step 2. Write the equation with the factors :
    ( x + 1 ) ( 3 x - 1 ) = 0 ( x + 1 ) ( 3 x - 1 ) = 0
    (4)
  3. Step 3. Determine the two solutions :

    We have

    x + 1 = 0 x + 1 = 0
    (5)

    or

    3 x - 1 = 0 3 x - 1 = 0
    (6)

    Therefore, x=-1x=-1 or x=13x=13.

  4. Step 4. Check the solutions: We substitute the answers back into the original equation and for both answers we find that the equation is true.
  5. Step 5. Write the final answer :

    3x2+2x-1=03x2+2x-1=0

     
    for x=-1x=-1 or x=13x=13.

Sometimes an equation might not look like a quadratic at first glance but turns into one with a simple operation or two. Remember that you have to do the same operation on both sides of the equation for it to remain true.

You might need to do one (or a combination) of:

  • Multiply both sides: For example,
    ax+b=cxx(ax+b)=x(cx)ax2+bx=cax+b=cxx(ax+b)=x(cx)ax2+bx=c
    (7)
  • Invert both sides: This is raising both sides to the power of -1-1. For example,
    1ax2+bx=c(1ax2+bx)-1=(c)-1ax2+bx1=1cax2+bx=1c1ax2+bx=c(1ax2+bx)-1=(c)-1ax2+bx1=1cax2+bx=1c
    (8)
  • Square both sides: This is raising both sides to the power of 2. For example,
    ax2+bx=c(ax2+bx)2=c2ax2+bx=c2ax2+bx=c(ax2+bx)2=c2ax2+bx=c2
    (9)

You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,

1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 ( invert both sides ) a x 2 + b x 1 = 1 c a x 2 + b x = 1 c ( a x 2 + b x ) 2 = ( 1 c ) 2 ( square both sides ) a x 2 + b x = 1 c 2 1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 ( invert both sides ) a x 2 + b x 1 = 1 c a x 2 + b x = 1 c ( a x 2 + b x ) 2 = ( 1 c ) 2 ( square both sides ) a x 2 + b x = 1 c 2
(10)

Exercise 2: Solving Quadratic Equations

Solve for xx: x+2=xx+2=x

Solution
  1. Step 1. Square both sides of the equation :

    Both sides of the equation should be squared to remove the square root sign.

    x + 2 = x 2 x + 2 = x 2
    (11)
  2. Step 2. Write equation in the form ax2+bx+c=0ax2+bx+c=0 :
    x + 2 = x 2 ( subtract x 2 from both sides ) x + 2 - x 2 = 0 ( divide both sides by - 1 ) - x - 2 + x 2 = 0 x 2 - x + 2 = 0 x + 2 = x 2 ( subtract x 2 from both sides ) x + 2 - x 2 = 0 ( divide both sides by - 1 ) - x - 2 + x 2 = 0 x 2 - x + 2 = 0
    (12)
  3. Step 3. Factorise the quadratic :
    x 2 - x + 2 x 2 - x + 2
    (13)

    The factors of x2-x+2x2-x+2 are (x-2)(x+1)(x-2)(x+1).

  4. Step 4. Write the equation with the factors :
    ( x - 2 ) ( x + 1 ) = 0 ( x - 2 ) ( x + 1 ) = 0
    (14)
  5. Step 5. Determine the two solutions :

    We have

    x + 1 = 0 x + 1 = 0
    (15)

    or

    x - 2 = 0 x - 2 = 0
    (16)

    Therefore, x=-1x=-1 or x=2x=2.

  6. Step 6. Check whether solutions are valid :

    Substitute x=-1x=-1

     
    into the original equation x+2=xx+2=x:

    LHS = ( - 1 ) + 2 = 1 = 1 but RHS = ( - 1 ) LHS = ( - 1 ) + 2 = 1 = 1 but RHS = ( - 1 )
    (17)

    Therefore LHS ≠≠ RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.

    Therefore x≠-1x≠-1.

    Now substitute x=2x=2 into original equation x+2=xx+2=x:

    LHS = 2 + 2 = 4 = 2 and RHS = 2 LHS = 2 + 2 = 4 = 2 and RHS = 2
    (18)

    Therefore LHS = RHS

    Therefore x=2x=2 is the only valid solution

  7. Step 7. Write the final answer :

    x+2=xx+2=x for x=2x=2 only.

Exercise 3: Solving Quadratic Equations

Solve the equation: x2+3x-4=0x2+3x-4=0.

Solution
  1. Step 1. Check if the equation is in the form ax2+bx+c=0ax2+bx+c=0 :

    The equation is in the required form, with a=1a=1.

  2. Step 2. Factorise the quadratic :

    You need the factors of 1 and 4 so that the middle term is +3+3 So the factors are:

    ( x - 1 ) ( x + 4 ) ( x - 1 ) ( x + 4 )

  3. Step 3. Solve the quadratic equation :
    x 2 + 3 x - 4 = ( x - 1 ) ( x + 4 ) = 0 x 2 + 3 x - 4 = ( x - 1 ) ( x + 4 ) = 0
    (19)

    Therefore x=1x=1 or x=-4x=-4.

  4. Step 4. Check the solutions:
    12+3(1)-4=012+3(1)-4=0
    (20)
    (-4)2+3(-4)-4=0(-4)2+3(-4)-4=0
    (21)

    Both solutions are valid.

  5. Step 5. Write the final solution :

    Therefore the solutions are x=1x=1 or x=-4x=-4.

Exercise 4: Solving Quadratic Equations

Find the roots of the quadratic equation 0=-2x2+4x-20=-2x2+4x-2.

Solution
  1. Step 1. Determine whether the equation is in the form ax2+bx+c=0ax2+bx+c=0, with no common factors. :

    There is a common factor: -2. Therefore, divide both sides of the equation by -2.

    - 2 x 2 + 4 x - 2 = 0 x 2 - 2 x + 1 = 0 - 2 x 2 + 4 x - 2 = 0 x 2 - 2 x + 1 = 0
    (22)
  2. Step 2. Factorise x2-2x+1x2-2x+1 :

    The middle term is negative. Therefore, the factors are (x-1)(x-1)(x-1)(x-1)

    If we multiply out (x-1)(x-1)(x-1)(x-1), we get x2-2x+1x2-2x+1.

  3. Step 3. Solve the quadratic equation :
    x 2 - 2 x + 1 = ( x - 1 ) ( x - 1 ) = 0 x 2 - 2 x + 1 = ( x - 1 ) ( x - 1 ) = 0
    (23)

    In this case, the quadratic is a perfect square, so there is only one solution for xx: x=1x=1.

  4. Step 4. Check the solution:

    -2(1)2+4(1)-2=0-2(1)2+4(1)-2=0.

  5. Step 5. Write the final solution :

    The root of 0=-2x2+4x-20=-2x2+4x-2 is x=1x=1.

Solving Quadratic Equations

  1. Solve for xx: (3x+2)(3x-4)=0(3x+2)(3x-4)=0
     
    Click here for the solution
  2. Solve for xx: (5x-9)(x+6)=0(5x-9)(x+6)=0
     
    Click here for the solution
  3. Solve for xx: (2x+3)(2x-3)=0(2x+3)(2x-3)=0
     
    Click here for the solution
  4. Solve for xx: (2x+1)(2x-9)=0(2x+1)(2x-9)=0
     
    Click here for the solution
  5. Solve for xx: (2x-3)(2x-3)=0(2x-3)(2x-3)=0
     
    Click here for the solution
  6. Solve for xx: 20x+25x2=020x+25x2=0
     
    Click here for the solution
  7. Solve for xx: 4x2-17x-77=04x2-17x-77=0
     
    Click here for the solution
  8. Solve for xx: 2x2-5x-12=02x2-5x-12=0
     
    Click here for the solution
  9. Solve for xx: -75x2+290x-240=0-75x2+290x-240=0
     
    Click here for the solution
  10. Solve for xx: 2x=13x2-3x+14232x=13x2-3x+1423
     
    Click here for the solution
  11. Solve for xx: x2-4x=-4x2-4x=-4
     
    Click here for the solution
  12. Solve for xx: -x2+4x-6=4x2-5x+3-x2+4x-6=4x2-5x+3
     
    Click here for the solution
  13. Solve for xx: x2=3xx2=3x
     
    Click here for the solution
  14. Solve for xx: 3x2+10x-25=03x2+10x-25=0
     
    Click here for the solution
  15. Solve for xx: x2-x+3x2-x+3
     
     
    Click here for the solution
  16. Solve for xx: x2-4x+4=0x2-4x+4=0
     
    Click here for the solution
  17. Solve for xx: x2-6x=7x2-6x=7
     
    Click here for the solution
  18. Solve for xx: 14x2+5x=614x2+5x=6
     
    Click here for the solution
  19. Solve for xx: 2x2-2x=122x2-2x=12
     
    Click here for the solution
  20. Solve for xx: 3x2+2y-6=x2-x+23x2+2y-6=x2-x+2
     
    Click here for the solution

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