solve 14th sum.........
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Here alpha = a and beta = b. Because it is difficult to write alpha and beta always.
Given a and b are roots of x^2 - 6x + k.
We know that sum of the roots = -b/a
= -(-6)/1
= 6.
a + b = 6 --------------- (1)
We know that product of roots = c/a
= k/1
= k.
ab = k ----------- (2)
Given that 3a - 2b = 20 ------------ (3)
On solving (1) * 3 & (3), we get
3a + 3b = 18
3a - 2b = 20
------------------
5b = -2
b = -2/5.
Substitute b = -2/5 in (1), we get
a + b = 6
a - 2/5 = 6
a = 6 + 2/5
a = 32/5.
Substitute a and b in (2), we get
ab = k
32/5 * -2/5 = k
-64/25 = k
Therefore the value of k = -64/25.
Hope this helps! -------------- Good Luck.
Given a and b are roots of x^2 - 6x + k.
We know that sum of the roots = -b/a
= -(-6)/1
= 6.
a + b = 6 --------------- (1)
We know that product of roots = c/a
= k/1
= k.
ab = k ----------- (2)
Given that 3a - 2b = 20 ------------ (3)
On solving (1) * 3 & (3), we get
3a + 3b = 18
3a - 2b = 20
------------------
5b = -2
b = -2/5.
Substitute b = -2/5 in (1), we get
a + b = 6
a - 2/5 = 6
a = 6 + 2/5
a = 32/5.
Substitute a and b in (2), we get
ab = k
32/5 * -2/5 = k
-64/25 = k
Therefore the value of k = -64/25.
Hope this helps! -------------- Good Luck.
siddhartharao77:
If possible brainliest it
thanxx.answer is -16
If the question is 3alpha + 2beta=20 then the answer will be -16.
ok
Nice
Answered by
1
alpha
= a and beta = b
---------- a and b are roots of x^2 - 6x + k. -----------------
********************** sum of the roots = -b/a *************************
= -(-6)/1
= 6.
a + b = 6 ____________ ( 1 )
product of roots = c/a
= k/1
= k.
ab = k ----------- (2)
---------------------- 3a - 2b = 20 ------------ (3)
(1) * 3 & (3), we get
3a + 3b = 18
3a - 2b = 20
------------------
5b = -2
b = -2/5.
Substitute b = -2/5 in (1),
a + b = 6
a - 2/5 = 6
a = 6 + 2/5
a = 32/5.
Substitute a and b in (2), we get
ab = k
32/5 * -2/5 = k
-64/25 = k
Therefore the value of k = -64/25.
= a and beta = b
---------- a and b are roots of x^2 - 6x + k. -----------------
********************** sum of the roots = -b/a *************************
= -(-6)/1
= 6.
a + b = 6 ____________ ( 1 )
product of roots = c/a
= k/1
= k.
ab = k ----------- (2)
---------------------- 3a - 2b = 20 ------------ (3)
(1) * 3 & (3), we get
3a + 3b = 18
3a - 2b = 20
------------------
5b = -2
b = -2/5.
Substitute b = -2/5 in (1),
a + b = 6
a - 2/5 = 6
a = 6 + 2/5
a = 32/5.
Substitute a and b in (2), we get
ab = k
32/5 * -2/5 = k
-64/25 = k
Therefore the value of k = -64/25.
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