Page 20 - mathsvol1ch1to3ans

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2
Simplifying the equation we getkx − (5 + 2k)x + k + 7
(5 + 2k)
Sum of the roots = 3α =
k
(k + 7)
2
Product of the roots = 2(α) =
k
2
5 + 2k (k + 7)
2 =
3k k
2
2(5 + 2k) = 9k(k + 7)
2
2
2(25 + 4k + 20k) = 9k + 63k
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2
2
50 + 8k + 40k = 9k + 63k
2
k + 23k − 50 = 0
(k − 2)(k + 25) = 0
The values of k are = 2 or − 25
2
5. If the difference of the roots of the equation 2x − (a + 1)x + a − 1 = 0 is equal to their product,
then prove that a = 2.
Solution:
a + 1
Sum of the roots =
2
a − 1
Difference of the roots = Product of the roots =
2
sum of roots + Difference of roots a
One of the root = =
2 2
Substituting this value in the given equation, we get
a a
2
2( ) − (a + 1) + a − 1 = 0
2 2
a 2 (a)(a + 1)
− + a − 1 = 0
2 2
2
a − a(a + 1) + 2a − 2 = 0
2
2
a − a − a + 2a − 2 = 0
a = 2
2
6. Find the condition that one of the roots of ax + bx + c may be (i) negative of the other, (ii)
thrice the other, (iii) reciprocal of the other.
Solution:Let one of the roots be α
(i) The roots are α and −α. The sum of the roots is 0. Hence b = 0.
b
(ii) The roots are α and 3α. Sum of the roots = 4α = - .
c a
2
Product of the roots are 3(α) =
a
2
b c
Hence, 3 − =
4a a
2
3b = 16ac
1 c
(iii) The roots are α and . Product of the roots = 1 = .
α a
Hence c = a.
2
2
7. If the equations x − ax + b = 0 and x − ex + f = 0 have one root in common and if the second
equation has equal roots, then prove that ae = 2(b + f).