What are the solutions to the equation (x^2)/(14)-x/7 =1 ?

The solutions to the equation (x^2)/(14)-x/7 =1 are x=1+sqrt(15),x=1-sqrt(15)

Find the zeros of (x^2)/(14)-x/7 =1

The zeros of (x^2)/(14)-x/7 =1 are x=1+sqrt(15),x=1-sqrt(15)

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\frac{x ^{2}}{14} - \frac{x}{7} = 1

x = 1 + 15, x = 1 - 15

Decimal Notation

x = 4.87298 \dots, x = - 2.87298 \dots

Solution steps

\frac{x ^{2}}{14} - \frac{x}{7} = 1

Find Least Common Multiplier of

14, 7 : 14

Multiply by LCM=

14

\frac{x ^{2}}{14} \cdot 14 - \frac{x}{7} \cdot 14 = 1 \cdot 14

Simplify

x^{2} - 2 x = 14

Move

14

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 14) = 215

x_{1, 2} = \frac{- ( - 2 ) \pm 2 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 15}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 15}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 15}{2 \cdot 1} : 1 + 15

x = \frac{- ( - 2 ) - 2 15}{2 \cdot 1} : 1 - 15

The solutions to the quadratic equation are:

x = 1 + 15, x = 1 - 15

2 x + 1 \leq 5

x^{4} - 5 x^{3} + 4 x^{2} - 16 x^{2} + 80 x - 64

(x^{3})^{\frac{8}{15}}

x^{4} + 4 x^{3} - 9 x^{2} - 16 x + 20

- x^{2} + 8 x - 12

What are the solutions to the equation (x^2)/(14)-x/7 =1 ?
The solutions to the equation (x^2)/(14)-x/7 =1 are x=1+sqrt(15),x=1-sqrt(15)
Find the zeros of (x^2)/(14)-x/7 =1
The zeros of (x^2)/(14)-x/7 =1 are x=1+sqrt(15),x=1-sqrt(15)