Determine the value of K for which the given value is the solution of the equation. 3x²+Kx+5=0;X=-5/3
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Answered by
0
ANSWER
2x
2
+kx−5=0
x
2
−3x−4=0----- Given
(c
1
a
2
−c
2
a
1
)
2
=(b
1
c
2
−b
2
c
1
)(a
1
b
2
−a
2
b
1
)
⇒[−5×1+4×2]
2
=[K(−4)+3(−5)][2(−3)−(1)(K)]
∴3
2
=[−4K−15][−6−K]
∴3
2
=(4K+15)(65K)
∴3
2
=(4K+15)(64K
)
∴9=24K+4K2+90+15K
∴0=4K
2
+39K+81
∴K=−3,−
4
27
Alternate solution
x
2
−3x−4=0
x
2
−4x+x−4=0
∴(x−4)(x+1)=0
∴x=4,−1
By keeping x=4
we get, 2(4)
2
+k×4−5=0
4k=−27
∴k=
4
−27
by keeping x=−1
we get, ∴2(−1)
2
+k(−1)−5=0
∴k=−3.
Answered by
0
Answer:
3x'2+kx+5=0
3×-5/3+k×-5/3+5=0
(-5-5k/3+5=0)×3
-15-5k+15=0
-5k=0
k=-0/5
k=0
Step-by-step explanation:
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