Math, asked by Anonymous, 7 months ago

Determine the value of K for which the given value is the solution of the equation. 3x²+Kx+5=0;X=-5/3
Pls help​

Answers

Answered by akhandpratapsingh92
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 abhishekmishra2175
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:

your answer friend

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