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Algebra & Inequalities | The Maths Tailor — UK Methods

Algebra & Inequalities

10 methods · 32 questions

The instant a prompt asks for which parameter values give 'distinct / equal / no real roots', 'positive (or negative) for all real $x$ ', or 'the line meets / does not meet the curve', do not solve the quadratic: impose the sign of its discriminant $b^{2} - 4 a c$ and read off the parameter range. Distinct real roots and 'graphs cross or touch' need $b^{2} - 4 a c \geq 0$ (strict $>$ for two distinct, $= 0$ for a repeated root); 'no real roots' or 'no intersection' needs $b^{2} - 4 a c < 0$ ; 'positive for all $x$ ' needs a positive leading coefficient together with $b^{2} - 4 a c < 0$ .

Trigger: a parameter sits inside a quadratic and the question is about how many real solutions exist or whether the quadratic keeps one sign.

Instances:
(i) reduce the curve-meets-line or two-equations system to one quadratic in $x$ , then apply the discriminant to the parameter;
(ii) for 'positive for all $x$ ', demand leading coefficient $> 0$ and discriminant $< 0$ ;
(iii) the discriminant is itself a quadratic in the parameter, so factorise it and solve that inequality to get the final range.

Linked questions (6)

TMUA 2018 Paper 1 — Q4

Consider the simultaneous equations

3 x^{2} + 2 x y = 4

x + y = a

where

a

is a real constant. Find the complete set of values of

a

for which the equations have two distinct real solutions for

x

.

(A) There are no values of

a

.
(B)

- 2 < a < 2

(C)

- 1 < a < 1

(D)

a = 0

(E)

a < - 1

a > 1

(F)

a < - 2

a > 2

(G) All real values of

a

Voir la correction

From the second equation,

y = a - x

. Substituting into the first:

3 x^{2} + 2 x (a - x) = 4 ⟹ 3 x^{2} + 2 a x - 2 x^{2} = 4 ⟹ x^{2} + 2 a x - 4 = 0.

The discriminant is

(2 a)^{2} - 4 (1) (- 4) = 4 a^{2} + 16

. Since

4 a^{2} \geq 0

for all real

a

, the discriminant is always strictly positive (

\geq 16 > 0

). Therefore the quadratic always has two distinct real solutions for

x

, for all real values of

a

. Hence the answer is G.

TMUA 2019 Paper 1 — Q2

Find the complete set of values of the real constant

k

for which the expression

x^{2} + k x + 2 x + 1 - 2 k

is positive for all real values of

x

.

(A)

- 12 < k < 0

(B)

k < - 12

k > 0

(C)

- 6 - 3 < k < 6 - 3

(D)

k < - 6 - 3

k > 6 - 3

(E)

- 2 < k < \frac{1}{2}

(F)

k < - 2

k > \frac{1}{2}

(G)

0 < k < 4

(H)

k < 0

k > 4

Voir la correction

Rewrite as

x^{2} + (k + 2) x + (1 - 2 k)

. For this quadratic (with positive leading coefficient) to be positive for all real

x

, the discriminant must be negative:

(k + 2)^{2} - 4 (1 - 2 k) < 0

. Expanding:

k^{2} + 4 k + 4 - 4 + 8 k < 0

, so

k^{2} + 12 k < 0

, i.e.

k (k + 12) < 0

. This holds when

- 12 < k < 0

. Hence the answer is A.

TMUA 2020 Paper 2 — Q1

Find the complete set of values of

k

for which the line

y = x - 2

crosses or touches the curve

y = x^{2} + k x + 2

.

(A)

- 1 \leq k \leq 3

(B)

- 3 \leq k \leq 5

(C)

- 4 \leq k \leq 4

(D)

k \leq - 1

k \geq 3

(E)

k \leq - 3

k \geq 5

(F)

k \leq - 4

k \geq 4

Voir la correction

Set the line and curve equal:

x - 2 = x^{2} + k x + 2

, giving

x^{2} + (k - 1) x + 4 = 0

. The line meets the curve when the discriminant is

\geq 0

(k - 1)^{2} - 16 \geq 0.

Expanding:

k^{2} - 2 k - 15 \geq 0

, which factorises as

(k - 5) (k + 3) \geq 0

. This holds when

k \leq - 3

k \geq 5

. Hence the answer is E.

TMUA 2023 Paper 1 — Q2

The graphs of

y = x^{2} + 5 x + 6

and

y = m x - 3

, where

m

is a constant, are plotted on the same set of axes.

Given that the graphs do not meet, what is the complete range of possible values of

m

?

(A)

- 1 < m < 11

(B)

m < - 1, m > 11

(C)

- 11 < m < 11

(D)

m < - 11, m > 11

(E)

- 11 < m < 1

(F)

m < - 11, m > 1

Voir la correction

If the graphs do not meet, the equation

x^{2} + 5 x + 6 = m x - 3

has no real solutions, i.e.

x^{2} + (5 - m) x + 9 = 0

has no real solutions. This requires the discriminant to be negative:

(5 - m)^{2} - 4 \times 1 \times 9 < 0

m^{2} - 10 m + 25 - 36 < 0

m^{2} - 10 m - 11 < 0

(m - 11) (m + 1) < 0.

Since the coefficient of

m^{2}

is positive, this quadratic in

m

is negative between its roots

m = - 1

and

m = 11

. So the complete range is

- 1 < m < 11

. Hence the answer is A.

TMUA 2023 Paper 2 — Q18

The equation

x^{4} + b x^{2} + c = 0

has four distinct real roots if and only if which of the following conditions is satisfied?

(A)

b^{2} > 4 c

(B)

b^{2} < 4 c

(C)

c > 0

and

b > 2 c

(D)

c > 0

and

b < - 2 c

(E)

c < 0

and

b < 0

(F)

c < 0

and

b > 0

Voir la correction

Substitute

u = x^{2}

: the equation becomes

u^{2} + b u + c = 0

. For four distinct real roots in

x

we need two distinct positive real roots

u_{1}, u_{2}

u

. The quadratic in

u

has two distinct real roots iff

b^{2} - 4 c > 0

. The roots are

u = \frac{- b \pm b ^{2} - 4 c}{2}

. For both roots to be positive: their product

c > 0

and their sum

- b > 0

, so

b < 0

. With

b < 0

and

c > 0

, both roots are positive automatically. We also need them distinct:

b^{2} - 4 c > 0

. Since

c > 0

b^{2} > 4 c

with

b < 0

means

b < - 2 c

. Combining: the necessary and sufficient condition is

c > 0

and

b < - 2 c

. Hence the answer is D.

TMUA Specimen Paper 1 — Q17

For what values of the non-zero real number

a

does the quadratic equation

a x^{2} + (a - 2) x = 2

have real distinct roots?

(A) All values of

a

(B)

a = - 2

(C)

a > - 2

(D)

a \neq = - 2

(E) No values of

a

Voir la correction

Rewrite as

a x^{2} + (a - 2) x - 2 = 0

. The discriminant is

(a - 2)^{2} - 4 \cdot a \cdot (- 2) = a^{2} - 4 a + 4 + 8 a = a^{2} + 4 a + 4 = (a + 2)^{2}

. The equation has real distinct roots when the discriminant is strictly positive, i.e.

(a + 2)^{2} > 0

. Since

(a + 2)^{2} \geq 0

for all real

a

and equals zero only when

a = - 2

, the condition is

a \neq = - 2

. (Note

a = 0

is excluded by the problem statement, but

(a + 2)^{2} > 0

already holds for all

a \neq = - 2

including

a \neq = 0

.) Hence the answer is D.