1.Algebra Booster

Quadratic Equations and Expressions 2.43
189. Now, x 2 – 9 £ 0
fi (x + 3)(x – 3) £ 0
fi –3 £ x £ 3 …(i)
Also, x 2 – 1 ≥ 0
fi (x + 1)(x – 1) ≥ 0
fi x £ –1, x ≥ 1
fi x Œ (– , –1] » [1, ) …(ii)
From Eqs (i) and (ii), we get
x Œ [–3, –1] » [1, 3]
190. Now, x 3 – 9x ≥ 0
fi x(x 2 – 9) ≥ 0
fi x(x + 3)(x – 3) ≥ 0
fi –3 £ x £ 0, x ≥ 3 …(i)
Also, x 3 + 4x £ 0
fi x(x 2 + 4) £ 0
fi x £ 0 …(ii)
From Eqs (i) and (ii), we get
x Œ [–3, 0]
198. We have,
|x| 3 – 3|x| + 2 = 0
fi (|x| – 1)(|x| – 2) = 0
fi (|x| – 1) = 0, (|x| – 2) = 0
fi |x| = 1, |x| = 2
fi x = ±1, x = ±2
199. We have,
x 2 – 4|x| + 3 = 0
fi |x| 3 – 4|x| + 3 = 0
fi (|x| – 1)(|x| – 3) = 0
fi (|x| – 1) = 0, (|x| – 3) = 0
fi |x| = 1, |x| = 3
fi x = ±1, x = ±3
Hence the sum of the roots = 1 – 1 + 3 – 3 = 0
200. We have,
|3x – 1| = |x + 5|
fi (3x – 1) = ±(x + 5)
Taking positive sign, we get
3x – 1 = x + 5
fi 2x = 6
fi x = 3
Taking negative sign, we get
3x – 1 = –x – 5
fi 4x = –4
fi x = –1
Hence, the solutions are x = –1, 3.
201. We have,
|2x – 5| = x – 3
fi (x 2 – x – 6) = ±(x + 2)
Taking positive sign, we get
2x – 5 = x – 3
fi 2x – x = 5 – 3
fi x = 2
Taking negative sign, we get
2x – 5 = –x + 3
fi 2x + x = 5 + 3
fi 3x = 8
fi x = 8/3
Hence, the solution set is {2, 8/3}.
202. We have,
|x 2 – x – 6| = x + 2
fi (x 2 – x – 6) = ±(x + 2)
Taking positive sign, we get
(x 2 – x – 6) = (x + 2)
fi (x 2 – 2x – 8) = 0
fi (x – 4)(x + 2) = 0
fi x = –2, 4.
Taking negative sign, we get
(x 2 – x – 6) = –(x + 2)
fi (x 2 – 4) = 0
fi x = ±2
Hence the solution set is {–2, 2, 4}.
203. We have 2|x – 2| + 3|x – 4| = 3
Case I: When x < 2
fi –2(x – 2) – 3(x – 4) = 3
fi –2x + 4 – 3x + 12 = 3
fi –5x = 3 – 16 = –13
fi x = 13/5
It is rejected, since x = 13/5 > 2.
Case II: When 2 £ x £ 4
fi 2(x – 2) – 3(x – 4) = 3
fi 2x – 4 – 3x + 12 = 3
fi –x = 3 – 8
fi x = 5
It is rejected, since x = 5 not lies in the given interval.
Case III: When x ≥ 4
fi 2(x – 2) + 3(x – 4) = 3
fi 2x – 4 + 3x – 12 = 3
fi 5x = 3 + 4 + 12 = 19
fi x = 19/5
It is rejected, since x = 19/5 < 4.
Hence, the solution set is f.
204. We have |x| + |x – 2| = 4
Case I: When x < 0.
fi –x – (x – 2) = 4
fi –x – x + 2 = 4
fi –2x = 4 – 2 = 2
fi x = –1
Case II: When 0 £ x £ 2
fi x – (x – 2) = 4
fi 2 = 4
It is not possible.
Case III: When x ≥ 2
fi x + x – 2 = 4
fi 2x = 4 + 2
fi 2x = 6
fi x = 3
Hence, the solution set is {–1, 3}
205. We have,
|x – 1| + |x – 3| = 2
fi |x – 1| + |3 – x| = 2
fi (x – 1)(3 – x) ≥ 0
fi (x – 1)(x – 3) £ 0
fi 1 £ x £ 3
fi x Π[1, 3]
206. We have,
|x 2 – 1| + |x 2 – 4| = 3