If the polynomial f(x) = x3 - kx2 + x + 6 and g (x) = x3 - kx2 + x + 6 be such that f(a) = 0 but g (a) unequal sign 0 , then (x - a) is a factor of
Answers
Step-by-step explanation:
tion 1:
Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
(i) f(x) = x2 − 2x − 8
(ii) g(s) = 4s2 − 4s + 1
(iii) h(t) = t2 − 15
(iv) 6x2 − 3 − 7x
(v) p(x)=x2+2
√
2
x-6
(vi) q(x)=
√
3
x2+10x+7
√
3
(vii)f(x)=x2-(
√
3
+1) x+
√
3
(viii) g(x) = a(x2 + 1) − x(a2 + 1)
(ix) h(s)=2s2-(1+2
√
2
)s+
√
2
(x) f(v)=v2+4
√
3
v-15
(xi) p(y)=y2+
3
√
5
2
y-5
(xii) q(y)=7y2-
11
3
y-
2
3
ANSWER:
(i) We have,
f(x) = x2 − 2x − 8
f(x) = x2 + 2x − 4x − 8
f(x) = x (x + 2) − 4(x + 2)
f(x) = (x + 2) (x − 4)
The zeros of f(x) are given by
f(x) = 0
x2 − 2x − 8 = 0
(x + 2) (x − 4) = 0
x + 2 = 0
x = −2
Or
x − 4 = 0
x = 4
Thus, the zeros of f(x) = x2 − 2x − 8 are α = −2 and β = 4
Now,
and
Therefore, sum of the zeros =
Product of the zeros
= − 2 × 4
= −8
and
Therefore,
Product of the zeros =
Hence, the relation-ship between the zeros and coefficient are verified.
(ii) Given
When have,
g(s) = 4s2 − 4s + 1
g(s) = 4s2 − 2s − 2s + 1
g(s) = 2s (2s − 1) − 1(2s − 1)
g(s) = (2s − 1) (2s − 1)
The zeros of g(s) are given by
Or
Thus, the zeros ofare
and
Now, sum of the zeros
and
Therefore, sum of the zeros =
Product of the zeros
and =
Therefore, the product of the zeros =
Hence, the relation-ship between the zeros and coefficient are verified.
(iii) Given
We have,
h(t) = t2 - 15h(t) = (t)2 - (
√
15
)2h(t) = (t +
√
15
) (t -
√
15
)
The zeros of are given by
h(t) = 0(t -
√
15
) (t +
√
15
) = 0(t -
√
15
) = 0t =
√
15
or (t +
√
15
) = 0t = -
√
15
Hence, the zeros of h(t) are α =
√
15
and β = -
√
15
.
Now,
Sum of the zeros
and =