What are the solutions to the equation (x-3)(x-5)=1 ?

The solutions to the equation (x-3)(x-5)=1 are x=4+sqrt(2),x=4-sqrt(2)

Find the zeros of (x-3)(x-5)=1

The zeros of (x-3)(x-5)=1 are x=4+sqrt(2),x=4-sqrt(2)

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(x - 3) (x - 5) = 1

x = 4 + 2, x = 4 - 2

Decimal Notation

x = 5.41421 \dots, x = 2.58578 \dots

Solution steps

(x - 3) (x - 5) = 1

Expand

(x - 3) (x - 5) : x^{2} - 8 x + 15

x^{2} - 8 x + 15 = 1

Move

1

to the left side

x^{2} - 8 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 14 = 22

x_{1, 2} = \frac{- ( - 8 ) \pm 2 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 2}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 2 2}{2 \cdot 1}

x = \frac{- ( - 8 ) + 2 2}{2 \cdot 1} : 4 + 2

x = \frac{- ( - 8 ) - 2 2}{2 \cdot 1} : 4 - 2

The solutions to the quadratic equation are:

x = 4 + 2, x = 4 - 2

(1.5 \cdot 1 0^{8}) x^{2} + x - 0.327 = 0, x

x^{2} = (32)^{2} + (32)^{2}

x^{2} - 16 + x (x - 4) = 0

(x - 1)^{2} + x^{2} = (x + 1)^{2}

x^{2} - 18 x - 2 = - 10

What are the solutions to the equation (x-3)(x-5)=1 ?
The solutions to the equation (x-3)(x-5)=1 are x=4+sqrt(2),x=4-sqrt(2)
Find the zeros of (x-3)(x-5)=1
The zeros of (x-3)(x-5)=1 are x=4+sqrt(2),x=4-sqrt(2)