A-Level Equations and inequalities: worked solution

Question

(a) Find the set of x with 3|x - 3| <= |1 - 2x|. (b) Without a calculator, solve (x + 18)/(x^2 + 5x - 14) >= -1.

(a) Solve 3|x − 3| <= |1 − 2x|. Step 1. Both sides are non-negative, since a modulus is never negative and 3 times a modulus is not either. When both sides of an inequality are non-negative, squaring keeps it the same way round, and squaring removes the modulus signs because |A|² = A². 9(x − 3)² <= (1 − 2x)² Step 2. Expand both sides. (x − 3)² = x² − 6x + 9, so 9(x − 3)² = 9x² − 54x + 81 (1 − 2x)² = 1 − 4x + 4x² = 4x² − 4x + 1 9x² − 54x + 81 <= 4x² − 4x + 1 Step 3. Bring everything to the left. 9x² − 54x + 81 − 4x² + 4x − 1 <= 0 5x² − 50x + 80 <= 0 Step 4. Divide by 5, which is positive so the sign does not change, then factorise. x² − 10x + 16 <= 0 Two numbers multiplying to 16 and adding to −10 are −2 and −8. (x − 2)(x − 8) <= 0 Step 5. A quadratic with a positive x² coefficient is <= 0 between (and at) its roots. 2 <= x <= 8 So the set is {x real : 2 <= x <= 8}. (b) Solve (x + 18)/(x² + 5x − 14) >= −1. Step 1. Do not multiply across by the denominator, whose sign is unknown. Add 1 to both sides instead. (x + 18)/(x² + 5x − 14) + 1 >= 0 Step 2. Factorise the denominator. Two numbers multiplying to −14 and adding to 5 are 7 and −2. x² + 5x − 14 = (x + 7)(x − 2) Step 3. Combine over that common denominator. [x + 18 + (x + 7)(x − 2)] / [(x + 7)(x − 2)] >= 0 numerator = x + 18 + x² + 5x − 14 = x² + 6x + 4 Step 4. The numerator has no nice factorisation, so find its roots with the formula (a = 1, b = 6, c = 4). x = (−6 ± sqrt(36 − 16))/2 = (−6 ± sqrt20)/2 sqrt20 = 2sqrt5, so x = (−6 ± 2sqrt5)/2 = −3 ± sqrt5 Hence x² + 6x + 4 = (x + 3 + sqrt5)(x + 3 − sqrt5), and the inequality is (x + 3 + sqrt5)(x + 3 − sqrt5) / [(x + 7)(x − 2)] >= 0 Step 5. List the critical values in increasing order and test one point in each of the five intervals. x = −7, x = −3 − sqrt5 (about −5.24), x = −3 + sqrt5 (about −0.76), x = 2. x = −8: numerator 64 − 48 + 4 = 20 > 0; denominator (−1)(−10) = 10 > 0; ratio > 0, so x < −7 works. x = −6: numerator 36 − 36 + 4 = 4 > 0; denominator (1)(−8) = −8 < 0; ratio < 0, fails. x = −3: numerator 9 − 18 + 4 = −5 < 0; denominator (4)(−5) = −20 < 0; ratio > 0, works. x = 0: numerator 4 > 0; denominator (7)(−2) = −14 < 0; ratio < 0, fails. x = 3: numerator 9 + 18 + 4 = 31 > 0; denominator (10)(1) = 10 > 0; ratio > 0, works. Step 6. Decide the endpoints. The numerator is zero at x = −3 − sqrt5 and x = −3 + sqrt5; zero satisfies >= 0, so both are included. The denominator is zero at x = −7 and x = 2, where the expression is undefined, so both are excluded. x < −7, or −3 − sqrt5 <= x <= −3 + sqrt5, or x > 2