What are the solutions to the equation 1/25 (x-4)^2-1/5 (x-6)^2=1 ?

The solutions to the equation 1/25 (x-4)^2-1/5 (x-6)^2=1 are x= 13/2-sqrt(5)i,x= 13/2+sqrt(5)i

Find the zeros of 1/25 (x-4)^2-1/5 (x-6)^2=1

The zeros of 1/25 (x-4)^2-1/5 (x-6)^2=1 are x= 13/2-sqrt(5)i,x= 13/2+sqrt(5)i

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\frac{1}{25} (x - 4)^{2} - \frac{1}{5} (x - 6)^{2} = 1

x = \frac{13}{2} - 5 i, x = \frac{13}{2} + 5 i

Solution steps

\frac{1}{25} (x - 4)^{2} - \frac{1}{5} (x - 6)^{2} = 1

Find Least Common Multiplier of

25, 5 : 25

Multiply by LCM=

25

\frac{1}{25} (x - 4)^{2} \cdot 25 - \frac{1}{5} (x - 6)^{2} \cdot 25 = 1 \cdot 25

Simplify

(x - 4)^{2} - 5 (x - 6)^{2} = 25

Expand

(x - 4)^{2} - 5 (x - 6)^{2} : - 4 x^{2} + 52 x - 164

- 4 x^{2} + 52 x - 164 = 25

Move

25

to the left side

- 4 x^{2} + 52 x - 189 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 52 \pm 5 2 ^{2} - 4 ( - 4 ) ( - 189 )}{2 ( - 4 )}

Simplify

5 2^{2} - 4 (- 4) (- 189) : 85 i

x_{1, 2} = \frac{- 52 \pm 8 5 i}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 52 + 8 5 i}{2 ( - 4 )}, x_{2} = \frac{- 52 - 8 5 i}{2 ( - 4 )}

x = \frac{- 52 + 8 5 i}{2 ( - 4 )} : \frac{13}{2} - 5 i

x = \frac{- 52 - 8 5 i}{2 ( - 4 )} : \frac{13}{2} + 5 i

The solutions to the quadratic equation are:

x = \frac{13}{2} - 5 i, x = \frac{13}{2} + 5 i

\frac{1}{\frac{5}{6}}

x^{2} + 6 x = - \frac{11}{4}

x^{2} - 3 x + 8 = 0

- \frac{1}{3} (1)

(4 + 2 i) - (7 - 3 i)

What are the solutions to the equation 1/25 (x-4)^2-1/5 (x-6)^2=1 ?
The solutions to the equation 1/25 (x-4)^2-1/5 (x-6)^2=1 are x= 13/2-sqrt(5)i,x= 13/2+sqrt(5)i
Find the zeros of 1/25 (x-4)^2-1/5 (x-6)^2=1
The zeros of 1/25 (x-4)^2-1/5 (x-6)^2=1 are x= 13/2-sqrt(5)i,x= 13/2+sqrt(5)i