Suppose k > 0. Let c and d be the roots of x² - kax - (k + 1)b = 0 and a, b be the roots of x² - kcx - (k + 1)d = 0, and a + b + c + d = 180, then k is equal to A. 11 B. 9 C. 7 D. 5
@@AkshayxV7 a + b = Kc - (i) c + d = ka - (ii) (i) + (ii) a + b + c + d = k(a + c) 180 = K(a + c) a + c = 180/k ->eqn 1 ab = - (k + 1) * d - (iii) cd = - (k + 1) * b - (iv) (iii)×(iv) abcd = (k+1)^2 bd ac = (k + 1) ^ 2 ->eqn2 → Put x = a in eqn x^2 -kcx -(k+1)d=0 a^2- kac - (k + 1)d = 0 - (v) → put x = c in eqn x^2 - kax -(k+1)b=0 c^2 - kac - (k + 1)b = 0 - (vi) (v) + (vi) [a^2 + c^2 - 2kac - (k + 1)(b + d) = 0] [main eqn.] → from eqn. a + b + c + d = k(a + c) (i) + (ii) b + d = k(a + c) - (a + c) b + d = (a + c)(k - 1) ->eqn3 Squaring Both side in eqn 1(because to find [a^2+c^2]for main eqn.) (a + c)^2 = (180/k)^2 a^2 + c^2 = (180/k)^2 - 2ac ->eqn 4 Substitute (b+d) and (a^2+c^2) from eqn 3 and 4 in main eqn (180/k)^2 - 2ac - 2kac -(k+1)(k-1)(a+c)=0 → Now Substitute (a+c) and (ac) from eqn 1 and 2 in above eqn (180/k)^2 -2(k + 1)^2 -2k(k + 1)^2 -(k^2 - 1)(180)/k) = 0 On simplifying, 2k^5 +6k^4 +186k^3 + 2k^2 -180k -(180)^2 = 0 → Now by using calculator put all four options value, which are value of K & you will see only 5 is satisfying the Eqn. So,5 is answer.
@@Rahul_Singh1729 what books I can try to get my sequence and series better or a book for overall improvement in all maths topic as im hell weak in mathematics
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Suppose k > 0. Let c and d be the roots of x² - kax - (k + 1)b = 0 and a, b be the roots of x² - kcx - (k + 1)d = 0, and a + b + c + d = 180, then k is equal to
A. 11
B. 9
C. 7
D. 5
5 ?
@@SanatR001 can you tell the solution please.
I don't know the answer.I am searching for it.
You have to find relarion between a and c
Just use sum and product
And also put c and a in the respective eqn @@yuvrajsingh-wk3py
B 9
@@AkshayxV7 a + b = Kc - (i)
c + d = ka - (ii)
(i) + (ii)
a + b + c + d = k(a + c)
180 = K(a + c)
a + c = 180/k ->eqn 1
ab = - (k + 1) * d - (iii)
cd = - (k + 1) * b - (iv)
(iii)×(iv)
abcd = (k+1)^2 bd
ac = (k + 1) ^ 2 ->eqn2
→ Put x = a in eqn x^2 -kcx -(k+1)d=0
a^2- kac - (k + 1)d = 0 - (v)
→ put x = c in eqn x^2 - kax -(k+1)b=0
c^2 - kac - (k + 1)b = 0 - (vi)
(v) + (vi)
[a^2 + c^2 - 2kac - (k + 1)(b + d) = 0] [main eqn.]
→ from eqn. a + b + c + d = k(a + c) (i) + (ii)
b + d = k(a + c) - (a + c)
b + d = (a + c)(k - 1) ->eqn3
Squaring Both side in eqn 1(because to find [a^2+c^2]for main eqn.)
(a + c)^2 = (180/k)^2
a^2 + c^2 = (180/k)^2 - 2ac ->eqn 4
Substitute (b+d) and (a^2+c^2) from eqn 3 and 4 in main eqn
(180/k)^2 - 2ac - 2kac -(k+1)(k-1)(a+c)=0
→ Now Substitute (a+c) and (ac) from eqn 1 and 2 in above eqn
(180/k)^2 -2(k + 1)^2 -2k(k + 1)^2 -(k^2 - 1)(180)/k) = 0
On simplifying,
2k^5 +6k^4 +186k^3 + 2k^2 -180k -(180)^2 = 0
→ Now by using calculator put all four options value, which are value of K & you will see only 5 is satisfying the Eqn.
So,5 is answer.
Sir sequence and series me miscellaneous series tough lagta hai sir
Try higher alegbra book. They have very good questions and you can learn alot from them.
@@Rahul_Singh1729 what books I can try to get my sequence and series better or a book for overall improvement in all maths topic as im hell weak in mathematics
complex nuber me geometry wala part
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Conjc section
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Conic sections
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