Home / IBDP Maths analysis and approaches Topic: SL 2.10 : Solving equations, both graphically and analytically HL Paper 1

# IBDP Maths analysis and approaches Topic: SL 2.10 : Solving equations, both graphically and analytically HL Paper 1

### Question

The graph of a polynomial function f of degree 4 is shown below.

Given that $${(x + {\text{i}}y)^2} = – 5 + 12{\text{i}},{\text{ }}x,{\text{ }}y \in \mathbb{R}$$ . Show that

(i)     $${x^2} – {y^2} = – 5$$ ;

(ii)     $$xy = 6$$ .[2]

A.a.

Hence find the two square roots of $$– 5 + 12{\text{i}}$$ .[5]

A.b.

For any complex number z , show that $${(z^*)^2} = ({z^2})^*$$ .[3]

A.c.

Hence write down the two square roots of $$– 5 – 12{\text{i}}$$ .[2]

A.d.

Explain why, of the four roots of the equation $$f(x) = 0$$ , two are real and two are complex.[2]

B.a.

The curve passes through the point $$( – 1,\, – 18)$$ . Find $$f(x)$$ in the form

$$f(x) = (x – a)(x – b)({x^2} + cx + d),{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}$$ .[5]

B.b.

Find the two complex roots of the equation $$f(x) = 0$$ in Cartesian form.[2]

B.c.

Draw the four roots on the complex plane (the Argand diagram).[2]

B.d.

Express each of the four roots of the equation in the form $$r{{\text{e}}^{{\text{i}}\theta }}$$ .[6]

B.e.

### Markscheme

(i)     $${(x + {\text{i}}y)^2} = – 5 + 12{\text{i}}$$

$${x^2} + 2{\text{i}}xy + {{\text{i}}^2}{y^2} = – 5 + 12{\text{i}}$$     A1

(ii)     equating real and imaginary parts     M1

$${x^2} – {y^2} = – 5$$     AG

$$xy = 6$$     AG

[2 marks]

A.a.

substituting     M1

EITHER

$${x^2} – \frac{{36}}{{{x^2}}} = – 5$$

$${x^4} + 5{x^2} – 36 = 0$$     A1

$${x^2} = 4,\, – 9$$     A1

$$x = \pm 2$$ and $$y = \pm 3$$     (A1)

OR

$$\frac{{36}}{{{y^2}}} – {y^2} = – 5$$

$${y^4} – 5{y^2} – 36 = 0$$     A1

$${y^2} = 9,\, – 4$$     A1

$${y^2} = \pm 3$$ and $$x = \pm 2$$     (A1)

Note: Accept solution by inspection if completely correct.

THEN

the square roots are $$(2 + 3{\text{i}})$$ and $$( – 2 – 3{\text{i}})$$     A1

[5 marks]

A.b.

EITHER

consider $$z = x + {\text{i}}y$$

$$z^* = x – {\text{i}}y$$

$${(z^*)^2} = {x^2} – {y^2} – 2{\text{i}}xy$$     A1

$$({z^2}) = {x^2} – {y^2} + 2{\text{i}}xy$$     A1

$$({z^2})^* = {x^2} – {y^2} – 2{\text{i}}xy$$     A1

$${(z^*)^2} = ({z^2})^*$$     AG

OR

$$z^* = r{{\text{e}}^{ – {\text{i}}\theta }}$$

$${(z^*)^2} = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}$$     A1

$${z^2} = {r^2}{{\text{e}}^{2{\text{i}}\theta }}$$     A1

$$({z^2})^* = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}$$     A1

$${(z^*)^2} = ({z^2})^*$$     AG

[3 marks]

A.c.

$$(2 – 3{\text{i}})$$ and $$( – 2 + 3{\text{i}})$$     A1A1

[2 marks]

A.d.

the graph crosses the x-axis twice, indicating two real roots     R1

since the quartic equation has four roots and only two are real, the other two roots must be complex     R1

[2 marks]

B.a.

$$f(x) = (x + 4)(x – 2)({x^2} + cx + d)$$     A1A1

$$f(0) = – 32 \Rightarrow d = 4$$     A1

Since the curve passes through $$( – 1,\, – 18)$$,

$$– 18 = 3 \times ( – 3)(5 – c)$$     M1

$$c = 3$$     A1

Hence $$f(x) = (x + 4)(x – 2)({x^2} + 3x + 4)$$

[5 marks]

B.b.

$$x = \frac{{ – 3 \pm \sqrt {9 – 16} }}{2}$$     (M1)

$$\Rightarrow x = – \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}$$     A1

[2 marks]

B.c.

A1A1

Note: Accept points or vectors on complex plane.

Award A1 for two real roots and A1 for two complex roots.

[2 marks]

B.d.

real roots are $$4{{\text{e}}^{{\text{i}}\pi }}$$ and $$2{{\text{e}}^{{\text{i}}0}}$$     A1A1

considering $$– \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}$$

$$r = \sqrt {\frac{9}{4} + \frac{7}{4}} = 2$$     A1

finding $$\theta$$ using $$\arctan \left( {\frac{{\sqrt 7 }}{3}} \right)$$     M1

$$\theta = \arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi {\text{ or }}\theta = \arctan \left( { – \frac{{\sqrt 7 }}{3}} \right) + \pi$$     A1

$$\Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi } \right)}}{\text{ or}} \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{ – \sqrt 7 }}{3}} \right) + \pi } \right)}}$$     A1

Note: Accept arguments in the range $$– \pi {\text{ to }}\pi {\text{ or }}0{\text{ to }}2\pi$$ .

[6 marks]

B.e.

### Question

The quadratic equation $${x^2} – 2kx + (k – 1) = 0$$ has roots $$\alpha$$ and $$\beta$$ such that $${\alpha ^2} + {\beta ^2} = 4$$. Without solving the equation, find the possible values of the real number $$k$$.

### Markscheme

$$\alpha + \beta = 2k$$    A1

$$\alpha \beta = k – 1$$    A1

$${(\alpha + \beta )^2} = 4{k^2} \Rightarrow {\alpha ^2} + {\beta ^2} + 2\underbrace {\alpha \beta }_{k – 1} = 4{k^2}$$    (M1)

$${\alpha ^2} + {\beta ^2} = 4{k^2} – 2k + 2$$

$${\alpha ^2} + {\beta ^2} = 4 \Rightarrow 4{k^2} – 2k – 2 = 0$$    A1

$$k = 1,{\text{ }} – \frac{1}{2}$$    A1

[6 marks]

### Question

Sketch the graphs of $$y = \frac{x}{2} + 1$$ and $$y = \left| {x – 2} \right|$$ on the following axes.

[3]

a.

Solve the equation $$\frac{x}{2} + 1 = \left| {x – 2} \right|$$.[4]

b.

### Markscheme

straight line graph with correct axis intercepts      A1

modulus graph: V shape in upper half plane      A1

modulus graph having correct vertex and y-intercept      A1

[3 marks]

a.

METHOD 1

attempt to solve $$\frac{x}{2} + 1 = x – 2$$     (M1)

$$x = 6$$      A1

Note: Accept $$x = 6$$ using the graph.

attempt to solve (algebraically) $$\frac{x}{2} + 1 = 2 – x$$     M1

$$x = \frac{2}{3}$$     A1

[4 marks]

METHOD 2

$${\left( {\frac{x}{2} + 1} \right)^2} = {\left( {x – 2} \right)^2}$$      M1

$$\frac{{{x^2}}}{4} + x + 1 = {x^2} – 4x + 4$$

$$0 = \frac{{3{x^2}}}{4} – 5x + 3$$

$$3{x^2} – 20x + 12 = 0$$

attempt to factorise (or equivalent)       M1

$$\left( {3x – 2} \right)\left( {x – 6} \right) = 0$$

$$x = \frac{2}{3}$$     A1

$$x = 6$$      A1

[4 marks]

b.
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