Question

If the roots of 4k^2-(5k+1)x + 5k = 0
differ by unity then the sum of all possible
values of k is​

Answer 1

✨✨Answer is here✨✨

➡️✨✨step the following and follow me please

α-β=1

(α-β)² =1²

α²+β²-2αβ=1 -:Equation 1

But (α+β)²= α²+β²+2αβ

Substituting (α+β)²-2αβ = α²+β²  in Equation 1

(α+β)²-2αβ-2αβ=1

(α+β)²-4αβ =1

A quadratic equation of the form ax²+bx+c=0 whose roots are α,β 

Then sum of roots α+β = -b/a

Product of roots = αβ = c/a 

So, 

(α+β)²-4αβ =1

(-b/a)²-4c/a=1 :-Equation 2

Comparing ax²+bx+c=0 with  4x²-(5k+1)x+5k

b= -(5k+1)   a=4   c=5k and substituting in Equation 2

(5k+1/4)²-4*5k/4=1

25k²+10k+1/16 - 5k =1

25k²+10k+1 =16(5k+1)

But 25k²+10k+1 =(5k+1)²

So,

(5k+1)² =16(5k+1)

5k+1 = 16

5k =16-1

5k=15

k = 3✨✨

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Answer 2

Step-by-step explanation:

  

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P(x) = 4x²+5k=(5k+1)x

= 4x²-(5k+1)x+5k

If α,β are roots of P(x)

According to question difference between roots is 1

α-β=1

(α-β)² =1²

α²+β²-2αβ=1-----Equation 1

But (α+β)²= α²+β²+2αβ

Substituting (α+β)²-2αβ = α²+β²  in Equation 1

(α+β)²-2αβ-2αβ=1

(α+β)²-4αβ =1

A quadratic equation of the form ax²+bx+c=0 whose roots are α,β 

Then sum of roots α+β = -b/a

Product of roots = αβ = c/a 

So, 

(α+β)²-4αβ =1

(-b/a)²-4c/a=1----Equation 2

Comparing ax²+bx+c=0 with  4x²-(5k+1)x+5k

b= -(5k+1)   a=4   c=5k and substituting in Equation 2

(5k+1/4)²-4*5k/4=1

25k²+10k+1/16 - 5k =1

25k²+10k+1 =16(5k+1)

But 25k²+10k+1 =(5k+1)²

So,

(5k+1)² =16(5k+1)

5k+1 = 16

5k =16-1

5k=15

k = 3