If the zeros of polynomial fx = 2x3-15x2+37x-30 are in ap. Find them.
Answers
Answered by
75
The answer is given below :
The given polynomial is
f(x) = 2x³ - 15x² + 37x - 30
Let us consider that the zeroes of the polynomial are (a - d), a and (a + d).
Then,
(a - d) + a + (a + d) = - (-15)/2
⇒ 3a = 15/2
⇒ a = 5/2 .....(i)
(a - d)a + (a - d)(a + d) + a(a + d) = 37/2
⇒ a² - ad + a² - d² + a² + ad = 37/2
⇒ 3a² - d² = 37/2
⇒ 3 (5/2)² - d² = 37/2, by (i)
⇒ 3 × (25/4) - d² = 37/2
⇒ d² = 75/4 - 37/2
⇒ d² = (75 - 74)/4
⇒ d² = 1/4
⇒ d = 1/2, (-1/2) .....(ii)
and
(a - d)a(a + d) = - (- 30/2)
⇒ a(a² - d²) = 15 .....(iii)
When, a = 5/2 and d = 1/2,
L.H.S of (iii)
= 5/2 × [(5/2)² - (1/2)²]
= 5/2 × (25/4 - 1/4)
= 5/2 × {(25 - 1)/4}
= 5/2 × 24/4
= 5/2 × 6
= 5 × 3
= 15
= R.H.S of (iii)
and the same is when d = -1/2
Now, the zeroes of the given polynomial are
(5/2 - 1/2), 5/2, (5/2 + 1/2), when d = 1/2
⇒ {(5 - 1)/2}, 5/2, {(5 + 1)/2}
⇒ 4/2, 5/2, 6/2
⇒ 2, 5/2, 3
When d = -1/2, the zeroes be
{5/2 - (-1/2)}, 5/2, {5/2 + (-1/2)}
⇒ (5/2 + 1/2), 5/2, (5/2 - 1/2)
⇒ {(5 + 1)/2}, 5/2, {(5 - 1)/2}
⇒ 6/2, 5/2, 4/2
⇒ 3, 5/2, 2
Therefore, the zeroes of the given polynomial are
2, 5/2 and 3.
Thank you for your question.
The given polynomial is
f(x) = 2x³ - 15x² + 37x - 30
Let us consider that the zeroes of the polynomial are (a - d), a and (a + d).
Then,
(a - d) + a + (a + d) = - (-15)/2
⇒ 3a = 15/2
⇒ a = 5/2 .....(i)
(a - d)a + (a - d)(a + d) + a(a + d) = 37/2
⇒ a² - ad + a² - d² + a² + ad = 37/2
⇒ 3a² - d² = 37/2
⇒ 3 (5/2)² - d² = 37/2, by (i)
⇒ 3 × (25/4) - d² = 37/2
⇒ d² = 75/4 - 37/2
⇒ d² = (75 - 74)/4
⇒ d² = 1/4
⇒ d = 1/2, (-1/2) .....(ii)
and
(a - d)a(a + d) = - (- 30/2)
⇒ a(a² - d²) = 15 .....(iii)
When, a = 5/2 and d = 1/2,
L.H.S of (iii)
= 5/2 × [(5/2)² - (1/2)²]
= 5/2 × (25/4 - 1/4)
= 5/2 × {(25 - 1)/4}
= 5/2 × 24/4
= 5/2 × 6
= 5 × 3
= 15
= R.H.S of (iii)
and the same is when d = -1/2
Now, the zeroes of the given polynomial are
(5/2 - 1/2), 5/2, (5/2 + 1/2), when d = 1/2
⇒ {(5 - 1)/2}, 5/2, {(5 + 1)/2}
⇒ 4/2, 5/2, 6/2
⇒ 2, 5/2, 3
When d = -1/2, the zeroes be
{5/2 - (-1/2)}, 5/2, {5/2 + (-1/2)}
⇒ (5/2 + 1/2), 5/2, (5/2 - 1/2)
⇒ {(5 + 1)/2}, 5/2, {(5 - 1)/2}
⇒ 6/2, 5/2, 4/2
⇒ 3, 5/2, 2
Therefore, the zeroes of the given polynomial are
2, 5/2 and 3.
Thank you for your question.
Answered by
30
For polynomial 2x³ - 15x² + 37x - 30
Le't assume that the zeroes of the polynomial are[a-d],a & [a+d]
Now, sum of Zeros = [-coefficient of x/coeffient of x³]
so we can write[a - d] + a + [a + d] = - [-15]/23a = 15/2
a = 5/2
and
[a - d]a + [a - d][a + d] + a[a + d] = 37/2a² - ad + a² - d² + a² + ad = 37/2
3a² - d² = 37/2 from eq 1 put value of a
3 [5/2]² - d² = 37/2
3 × [25/4] - d² = 37/2
d² = 75/4 - 37/2
d² = [75 - 74]/4
d² = 1/4
d = [-1/2], 1/2
also
[a - d]a[a + d] = - [- 30/2]a(a² - d²) = 15
put a = 5/2 & d = 1/2 we get
5/2 × [(5/2)² - (1/2)²]=155/2 × [25/4 - 1/4]=15
5/2 × {(25 - 1)/4}=15
5/2 × 24/4 = 15
5/2 × 6 = 15
5 × 3=15
15=15
similarly for d = -1/2
we ca say that zeros of the polynomial are
[5/2 - 1/2], 5/2, [5/2 + 1/2], when d = 1/2 and
{5/2 - (-1/2)}, 5/2, {5/2 + (-1/2)} when d=-1/2
so the zeros are2, 3 and 5/2
Le't assume that the zeroes of the polynomial are[a-d],a & [a+d]
Now, sum of Zeros = [-coefficient of x/coeffient of x³]
so we can write[a - d] + a + [a + d] = - [-15]/23a = 15/2
a = 5/2
and
[a - d]a + [a - d][a + d] + a[a + d] = 37/2a² - ad + a² - d² + a² + ad = 37/2
3a² - d² = 37/2 from eq 1 put value of a
3 [5/2]² - d² = 37/2
3 × [25/4] - d² = 37/2
d² = 75/4 - 37/2
d² = [75 - 74]/4
d² = 1/4
d = [-1/2], 1/2
also
[a - d]a[a + d] = - [- 30/2]a(a² - d²) = 15
put a = 5/2 & d = 1/2 we get
5/2 × [(5/2)² - (1/2)²]=155/2 × [25/4 - 1/4]=15
5/2 × {(25 - 1)/4}=15
5/2 × 24/4 = 15
5/2 × 6 = 15
5 × 3=15
15=15
similarly for d = -1/2
we ca say that zeros of the polynomial are
[5/2 - 1/2], 5/2, [5/2 + 1/2], when d = 1/2 and
{5/2 - (-1/2)}, 5/2, {5/2 + (-1/2)} when d=-1/2
so the zeros are2, 3 and 5/2
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