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

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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Complete the square for x^{2} + 10x + 24

We want to make x^{2} + 10x + 24 look like:

(x + a)^{2} – b, where a and b are numbers.

Make x^{2} + 10x → (x + 5)^{2} by halving the 10

(x + 5)^{2} → (x + 5)^{2} – 25 by removing the 5^{2}

So x^{2} + 10x = (x + 5)^{2} – 25 (check it)

and finally put back the 24

So x^{2} + 10x + 24 = (x + 5)^{2} – 1

To solve x^{2} + 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

Complete the square for 3x^{2} – 18x + 10

We want to make 3x^{2} – 18x + 10 look like:

a(x + b)^{2} + c, where a, b and c are numbers.

We begin by factoring out the coefficient of x^{2}, in this case 3:

3[x^{2} – 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 3^{2}

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 3x^{2} – 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

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 3x^{2} + 12x + 2 in the form a(x + b)^{2} + c where a, b, and c are integers

Factorise first: 3[x^{2} + 4x] + 2

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

Rearrange: 3(x + 2)^{2} – 12 + 2

So 3x^{2} + 12x + 2 is 3(x + 2)^{2} – 10

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

f(x) = 3x^{2} + 12x + 2 = 3(x + 2)^{2} – 10

This is a f(x) = x^{2} 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)**

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

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