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Inside Collection (Textbook):

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. 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.

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.

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

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

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: Text here
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)

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 to 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 to 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:

L H S = ( - 1 ) + 2 = 1 = 1 b u t R H S = ( - 1 ) L H S = ( - 1 ) + 2 = 1 = 1 b u t R H S = ( - 1 )
(17)

Therefore LHSRHS. 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:

L H S = 2 + 2 = 4 = 2 a n d R H S = 2 L H S = 2 + 2 = 4 = 2 a n d R H S = 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.

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. Write the final solution :

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

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
(20)
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
(21)

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

4. Step 4. Write the final solution :

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

1. Solve for xx: (3x+2)(3x-4)=0(3x+2)(3x-4)=0

2. Solve for aa: (5a-9)(a+6)=0(5a-9)(a+6)=0

3. Solve for xx: (2x+3)(2x-3)=0(2x+3)(2x-3)=0

4. Solve for xx: (2x+1)(2x-9)=0(2x+1)(2x-9)=0

5. Solve for xx: (2x-3)(2x-3)=0(2x-3)(2x-3)=0

6. Solve for xx: 20x+25x2=020x+25x2=0

7. Solve for aa: 4a2-17a-77=04a2-17a-77=0

8. Solve for xx: 2x2-5x-12=02x2-5x-12=0

9. Solve for bb: -75b2+290b-240=0-75b2+290b-240=0

10. Solve for yy: 2y=13y2-3y+14232y=13y2-3y+1423

11. Solve for θθ: θ2-4θ=-4θ2-4θ=-4

12. Solve for qq: -q2+4q-6=4q2-5q+3-q2+4q-6=4q2-5q+3

13. Solve for tt: t2=3tt2=3t

14. Solve for ww: 3w2+10w-25=03w2+10w-25=0

15. Solve for vv: v2-v+3v2-v+3


16. Solve for xx: x2-4x+4=0x2-4x+4=0

17. Solve for tt: t2-6t=7t2-6t=7

18. Solve for xx: 14x2+5x=614x2+5x=6

19. Solve for tt: 2t2-2t=122t2-2t=12

20. Solve for yy: 3y2+2y-6=y2-y+23y2+2y-6=y2-y+2


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