# IBDP Maths analysis and approaches Topic: AHL 2.12:Sum and product of the roots of polynomial equations HL Paper 1

### Question:  [Maximum mark: 4]

Consider the equation kx2 – (k + 3) x + 2k + 9 = 0, where k ∈ R.
(a) Write down an expression for the product of the roots, in terms of k .
(b) Hence or otherwise, determine the values of k such that the equation has one positive and one negative real root.

Ans:

(a) product of roots $$=\frac{2k+9}{k}$$

(b) recognition that the product of the roots will be negative
$$=\frac{2k+9}{k}$$ < 0

critical values $$k = 0, -\frac{9}{2}$$

$$-\frac{9}{2}<k<0$$

### Question

The cubic equation x3 – kx2 + 3k = 0 where k > 0 has roots α , β and α + β .

Given that α.β =$$\frac{-k^2}{4}$$ , find the value of k .[Maximum mark: 5]

Ans:

### Question

The equation $$5{x^3} + 48{x^2} + 100x + 2 = a$$ has roots $${r_1}$$, $${r_2}$$ and $${r_3}$$.

Given that $${r_1} + {r_2} + {r_3} + {r_1}{r_2}{r_3} = 0$$, find the value of a.

### Markscheme

$${r_1} + {r_2} + {r_3} = \frac{{ – 48}}{5}$$     (M1)(A1)

$${r_1}{r_2}{r_3} = \frac{{a – 2}}{5}$$     (M1)(A1)

$$\frac{{ – 48}}{5} + \frac{{a – 2}}{5} = 0$$     M1

$$a = 50$$     A1

Note:     Award M1A0M1A0M1A1 if answer of 50 is found using $$\frac{{48}}{5}$$ and $$\frac{{2 – a}}{5}$$.

[6 marks]

### Question

The roots of a quadratic equation $$2{x^2} + 4x – 1 = 0$$ are $$\alpha$$ and $$\beta$$.

Without solving the equation,

(a)     find the value of $${\alpha ^2} + {\beta ^2}$$;

(b)     find a quadratic equation with roots $${\alpha ^2}$$ and $${\beta ^2}$$.

### Markscheme

(a) using the formulae for the sum and product of roots:

$$\alpha + \beta = – 2$$     A1

$$\alpha \beta = – \frac{1}{2}$$     A1

$${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} – 2\alpha \beta$$     M1

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

$$= 5$$     A1

Note:     Award M0 for attempt to solve quadratic equation.

[4 marks]

(b)     $$(x – {\alpha ^2})(x – {\beta ^2}) = {x^2} – ({\alpha ^2} + {\beta ^2})x + {\alpha ^2}{\beta ^2}$$     M1

$${x^2} – 5x + {\left( { – \frac{1}{2}} \right)^2} = 0$$     A1

$${x^2} – 5x + \frac{1}{4} = 0$$

Note:     Final answer must be an equation. Accept alternative correct forms.

[2 marks]

Total [6 marks]

### Question

The quadratic equation $$2{x^2} – 8x + 1 = 0$$ has roots $$\alpha$$ and $$\beta$$.

Without solving the equation, find the value of

(i)     $$\alpha + \beta$$;

(ii)     $$\alpha \beta$$.[2]

a.

Another quadratic equation $${x^2} + px + q = 0,{\text{ }}p,{\text{ }}q \in \mathbb{Z}$$ has roots $$\frac{2}{\alpha }$$ and $$\frac{2}{\beta }$$.

Find the value of $$p$$ and the value of $$q$$.[4]

b.

## Markscheme

using the formulae for the sum and product of roots:

(i)     $$\alpha + \beta = 4$$     A1

(ii)     $$\alpha \beta = \frac{1}{2}$$     A1

Note:     Award A0A0 if the above results are obtained by solving the original equation (except for the purpose of checking).

[2 marks]

a.

METHOD 1

required quadratic is of the form $${x^2} – \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)x + \left( {\frac{2}{\alpha }} \right)\left( {\frac{2}{\beta }} \right)$$     (M1)

$$q = \frac{4}{{\alpha \beta }}$$

$$q = 8$$     A1

$$p = – \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)$$

$$= – \frac{{2(\alpha + \beta )}}{{\alpha \beta }}$$     M1

$$= – \frac{{2 \times 4}}{{\frac{1}{2}}}$$

$$p = – 16$$     A1

Note:     Accept the use of exact roots

METHOD 2

replacing $$x$$ with $$\frac{2}{x}$$     M1

$$2{\left( {\frac{2}{x}} \right)^2} – 8\left( {\frac{2}{x}} \right) + 1 = 0$$

$$\frac{8}{{{x^2}}} – \frac{{16}}{x} + 1 = 0$$     (A1)

$${x^2} – 16x + 8 = 0$$

$$p = – 16$$ and $$q = 8$$     A1A1

Note:     Award A1A0 for $${x^2} – 16x + 8 = 0$$ ie, if $$p = – 16$$ and $$q = 8$$ are not explicitly stated.

[4 marks]

Total [6 marks]

b.

### Question

The quadratic equation $$2{x^2} – 8x + 1 = 0$$ has roots $$\alpha$$ and $$\beta$$.

Without solving the equation, find the value of

(i)     $$\alpha + \beta$$;

(ii)     $$\alpha \beta$$.[2]

a.

Another quadratic equation $${x^2} + px + q = 0,{\text{ }}p,{\text{ }}q \in \mathbb{Z}$$ has roots $$\frac{2}{\alpha }$$ and $$\frac{2}{\beta }$$.

Find the value of $$p$$ and the value of $$q$$.[4]

b.

## Markscheme

using the formulae for the sum and product of roots:

(i)     $$\alpha + \beta = 4$$     A1

(ii)     $$\alpha \beta = \frac{1}{2}$$     A1

Note:     Award A0A0 if the above results are obtained by solving the original equation (except for the purpose of checking).

[2 marks]

a.

METHOD 1

required quadratic is of the form $${x^2} – \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)x + \left( {\frac{2}{\alpha }} \right)\left( {\frac{2}{\beta }} \right)$$     (M1)

$$q = \frac{4}{{\alpha \beta }}$$

$$q = 8$$     A1

$$p = – \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)$$

$$= – \frac{{2(\alpha + \beta )}}{{\alpha \beta }}$$     M1

$$= – \frac{{2 \times 4}}{{\frac{1}{2}}}$$

$$p = – 16$$     A1

Note:     Accept the use of exact roots

METHOD 2

replacing $$x$$ with $$\frac{2}{x}$$     M1

$$2{\left( {\frac{2}{x}} \right)^2} – 8\left( {\frac{2}{x}} \right) + 1 = 0$$

$$\frac{8}{{{x^2}}} – \frac{{16}}{x} + 1 = 0$$     (A1)

$${x^2} – 16x + 8 = 0$$

$$p = – 16$$ and $$q = 8$$     A1A1

Note:     Award A1A0 for $${x^2} – 16x + 8 = 0$$ ie, if $$p = – 16$$ and $$q = 8$$ are not explicitly stated.

[4 marks]

Total [6 marks]

b.
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