If one root of the the coordinator equation ke a x kx square k square+ 2x_8×0is_2 then what is the value of x
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The given quadratic equation is 8x
2
−6x+k=0, comparing it with ax
2
+bx+c.
⇒ Then, a=8,b=−6,c=k
⇒ It is given that one root of this equation is square of the other root. So, if we assume one root to be p, the other root can be p
2
. So, we assume the roots to be p and p
2
.
⇒ Sum of the roots =
a
−b
∴ p+p
2
=
8
−(−6)
∴ p+p
2
=
4
3
∴ 4p+4p
2
=3
∴ 4p
2
+4p−3=0 ----------- ( 1 )
⇒ Product of the roots =
a
c
∴ p×p
2
=
8
k
∴ p
3
=
8
k
------- ( 2 )
⇒ 4p
2
+4p−3=0 [ From ( 1 ) ]
⇒ 4p
2
−2p+6p−3=0
⇒ 2p(2p−1)+3(2p−1)=0
⇒ (2p+3)(2p−1)=0
∴ p=
2
−3
and p=
2
1
Now, putting p=
2
1
in equation ( 2 ) we get,
⇒ (
2
1
)
3
=
8
k
⇒
8
1
=
8
k
∴ k=1
Now, using p=
2
−3
in equation ( 2 )
⇒ (
2
−3
)
3
=
8
k
⇒
8
−27
=
8
k
∴ k=−27
∴ Values of k are 1 and −27
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