Exercise 1H
P1^
1
EXERCISE 1H 1 Solve the following inequalities.
(i) 5 a + 6 2 a + 24 (ii) 3 b − 5 b − 1
(iii) 4(c − 1) 3(c − 2) (iv) d − 3(d + 2) 2(1 + 2 d)
(v) 12 e+ 312 e (vi) −f − 2 f − 3 4(1 + f)
(vii) 5(2 − 3 g) + g 8(2g − 4) (viii) 3(h + 2) − 2(h − 4) 7(h + 2)
2 Solve the following inequalities by sketching the curves of the functions
involved.
(i) p^2 − 5 p + 4 < 0 (ii) p^2 − 5 p + 4 0
(iii) x^2 + 3x + 2 0 (iv) x^2 + 3 x − 2
(v) y^2 − 2 y − 3 0 (vi) z(z − 1) 20
(vii) q^2 − 4 q + 4 0 (viii) y(y − 2) 8
(ix) 3 x^2 + 5 x − 2 0 (x) 2 y^2 − 11 y − 6 0
(xi) 4 x − 3 x^2 (xii) 10 y^2 y + 3
3 Find the set of values of k for which each of these equations has two real roots.
(i) 2 x^2 − 3 x + k = 0 (ii) kx^2 + 4 x − 1 = 0
(iii) 5 x^2 + kx + 5 = 0 (iv) 3 x^2 + 2 kx + k = 0
4 Find the set of values of k for which each of these equations has no real roots.
(i) x^2 − 6 x + k = 0 (ii) kx^2 + x − 2 = 0
(iii) 4 x^2 − kx + 4 = 0 (iv) 2 kx^2 − kx + 1 = 0
KEY POINTS
1 The quadratic formula for solving ax^2 + bx + c = 0 is
x bb ac
a
=−± −
(^24)
2
where b^2 − 4 ac is called the discriminant.
If b^2 − 4 ac 0, the equation has two real roots.
If b^2 − 4 ac = 0, the equation has one repeated root.
If b^2 − 4 ac 0, the equation has no real roots.
2 To solve a pair of simultaneous equations where one equation is non-linear:
●●first make x or y the subject of the linear equation
●●then substitute this rearranged equation for x or y in the non-linear equation
●●solve to find y or x
●●substitute back into the linear equation to find pairs of solutions.
3 Linear inequalities are dealt with like equations but if you multiply or divide
by a negative number you must reverse the inequality sign.
4 When solving a quadratic inequality it is advisable to sketch the graph.