if the roots of 4x^2+5k=(5k+1)x differ by unity, then find the values of k.
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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
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Hope this helped you.................
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you.................
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