Page 23 - mathsvol1ch1to3ans

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2
2
Comparing coefficients of x we have 2a+b = −1 ⇒ b = −3. Hence the equation is x −3x−11.
√ !
3 ± 53
Solving for the roots we have, .
2
4
3. Find the real roots of x = 16.
4
2
Solution:Since x − 16 = (x + 4)(x + 2)(x − 2), the real roots are −2, 2.
2
2
4. Solve (2x + 1) − (3x + 2) = 0.
2
2
Solution: Since (3x + 2) > (2x + 1), we can rewrite as (3x + 2) − (2x + 1) = 0. Hence we
have [(3x + 2) + (2x + 1)] × [(3x + 2) − (2x + 1)] = (5x + 3)(x + 1). Hence values of x are
3
− , −1.
5
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Exercise - 2.12
4
1. Factorize: x + 1. (Hint: Try completing the square.)
Solution:
4
2 2
x + 1 = (x ) + 1
2
2
2 2
= (x ) + 2x − 2x + 1
2 2
2
= [(x ) + 2x + 1] − 2x 2
√
2
2
= (x + 1) − ( 2x) 2
√ √
2
2
= (x + 1 + 2x)(x + 1 − 2x)
√ √
2
2
= (x + 2x + 1)(x − 2x + 1)
3
2
2
2. If x + x + 1 is a factor of the polynomial 3x + 8x + 8x + a, then find the value of a.
2
2
3
Solution:Let 3x + 8x + 8x + a = (x + x + 1)(px + q)
3
2
By comparing the coefficients of x we have p = 3. By comparing the coefficients of x we have
3
2
2
p + q = 8 ⇒ q = 5. Hence 3x + 8x + 8x + a = (x + x + 1)(3x + 5). By comparing the
coefficients of constant term we have a = 5.
Exercise - 2.13
3
x (x − 1)
1. Find all values of x for which > 0.
(x − 2)
Solution: The critical points are 0,1,2.
3
x (x − 1)
Interval Sign of x Sign of (x − 1) Sign of (x − 2) Sign of > 0
(x − 2)
(−∞, 0) − − − −
(0, 1) + − − +
(1, 2) + + − −

(2, ∞) + + + +

So the inequality is satisfied in (0, 1) ∪ (2, ∞).