Algebra & Inequalities

11 methods · 38 questions

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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 (3)

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

AThere are no values of $a$ .
B $- 2 < a < 2$
C $- 1 < a < 1$
D $a = 0$
E $a < - 1$ or $a > 1$
F $a < - 2$ or $a > 2$
GAll real values of $a$

Worked solution

Answer:G

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 $k < - 12$ or $k > 0$
C $- 6 - 3 < k < 6 - 3$
D $k < - 6 - 3$ or $k > 6 - 3$
E $- 2 < k < \frac{1}{2}$
F $k < - 2$ or $k > \frac{1}{2}$
G $0 < k < 4$
H $k < 0$ or $k > 4$

Worked solution

Answer:A

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 1 · Q20

For how many values of

a

is the equation

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

satisfied by exactly two distinct values of

x

A $0$
B $1$
C $2$
D $3$
E $4$
Fmore than 4

Worked solution

Answer:C

The equation has solutions

x = a

(from the linear factor) and

x = \frac{1 \pm 1 - 4 a}{2}

(from the quadratic, when its discriminant

1 - 4 a \geq 0

). For exactly two distinct solutions, either: (Case 1) the quadratic has a repeated root (discriminant

= 0

) and that root is different from

a

: discriminant

= 0

gives

a = \frac{1}{4}

, and the repeated root is

\frac{1}{2} \neq = \frac{1}{4}

, so this works. Or (Case 2)

a

is itself a root of the quadratic

x^{2} - x + a = 0

: substituting

x = a

gives

a^{2} - a + a = 0

, so

a^{2} = 0

, giving

a = 0

. Check: when

a = 0

, the equation becomes

x (x^{2} - x) = x^{2} (x - 1) = 0

, giving exactly two distinct values

x = 0

and

x = 1

. So there are two values of

a

(

a = 0

and

a = \frac{1}{4}

) for which the equation has exactly two distinct solutions. Hence the answer is C.