Complete the Square, turning points

Year 10/11 maths: Find quadratics roots by completing the square, sketching, turning points




Complete Square MENU
Find roots, turning points graphically
quadra<p>If we plot a quadratic equation we get this type of graphtic graph

If you plot a quadratic equation you get a quadratic graph.

The Roots are the solutions to the equation.
You can solve quadratics equations graphically or by an algebraic method like completing the square.

Completing the square also provides information on how to sketch a quadratic equation


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Simple complete the square  Menu

Complete the square for x2 + 10x + 24

We want to make x2 + 10x + 24 look like:
(x + a)2 – b, where a and b are numbers.


Make x2 + 10x → (x + 5)2 by halving the 10
(x + 5)2 → (x + 5)2 – 25  by removing the 52


So x2 + 10x = (x + 5)2 – 25 (check it)
and finally put back the 24


So x2 + 10x + 24 = (x + 5)2 – 1


To solve x2 + 10x + 24 = 0
we solve (x + 5)2 – 1 = 0


Rearrange  (x + 5)2 = 1
square root  x + 5 = ± √1
rearrange so x = ± 1 – 5
x = – 4 or – 6

Complete the square and solve:

 x² – 12x + 5 = 0 give answer as a √ surd

ANS

Harder complete the square  Menu

Complete the square for 3x2 – 18x + 10

We want to make 3x2 – 18x + 10 look like:
a(x + b)2 + c, where a, b and c are numbers.


We begin by factoring out the coefficient of x2, in this case 3:
3[x2 – 6x] + 10 = 0.

The number in the complete square will be half the coefficient of x, i.e. 3
Remember to subtract 9 which is the 32

 3[(x – 3)2 – 9 ] + 10 = 0


Simplify by removing the 9 from within the square brackets but remember to multiply it.
You can remove the square brackets [ now

  3(x – 3)2 – 27 + 10 = 0 or 3(x – 3)2 – 17 = 0


Check that 3x2 – 18x + 10 = 3(x – 3)2 – 17 by substituting a value for x ie x = 1

Now solve 3(x – 3)2 – 17 = 0


Rearrange and divide (x – 3)2 = 17/3
square root  x – 3 = ± √(17/3)
rearrange so x = 3 ± √(17/3)

Q: Complete the square and solve:
 2x² – 12x + 7 = 0 give answer as a √ surd

 3x² – 9x – 13/4 = 0 give answer as a √ surd

ANS

Turning points  Menu

The turning point of a curved graph is the point where the gradient changes direction.

To find the turning point of a quadratic equation first simplify it by completing the square.

a) Write 3x2 + 12x + 2 in the form a(x + b)2 + c where a, b, and c are integers

Factorise first: 3[x2 + 4x] + 2

Complete the square: 3[(x + 2)2 – 4] + 2

Rearrange: 3(x + 2)2 – 12 + 2

So 3x2 + 12x + 2 is 3(x + 2)2 – 10


b) Hence, or otherwise, sketch the graph of f(x) = 3x2 + 12x + 2 showing the coordinates of the turning point and the coordinates of any intercepts with the axes.
 f(x) = 3x2 + 12x + 2 = 3(x + 2)2 – 10

This is a f(x) = x2 graph that has been transformed

Translate 2 to left x direction, stretch with SF3 in y then 10 down in y.
(You do it in this order and NOT stretch in y last which would pull 10 down to 30)

The turning point is the lowest value for f(x) when x = –2 to make the bracketed squared value zero.
f(x) = –10 , so turning point is (–2, –10)


turning points

Find the intersect on y axis when x = 0
so f(0) = 3(0 + 2)2 – 10, f(0) = 12 – 10 = 2

Find the intersect on x axis when f(x) = 0
so f(x) = 3(x + 2)2 – 10 = 0


Solve f(x) = 3(x + 2)2 – 10 = 0

 (x + 2)2 = 10/3

 (x + 2) = ± √(10/3)

x = –2 ± √(10/3)

Intersects are (0, 2) on y axis; (–2 + √(10/3), 0) on x axis; (–2 – √(10/3), 0) on x axis


Sketch the following curves showing their turning points and intersects with the x and y axis

 f(x)= 2x² – 12x + 7

 f(x)=3x² – 9x – 13/4

ANS


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