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

The solutions to the equation (x^2+1)/5-(x+1)/4 =1 are x=3,x=-7/4

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

The zeros of (x^2+1)/5-(x+1)/4 =1 are x=3,x=-7/4

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\frac{x ^{2} + 1}{5} - \frac{x + 1}{4} = 1

x = 3, x = - \frac{7}{4}

Decimal Notation

x = 3, x = - 1.75

Solution steps

\frac{x ^{2} + 1}{5} - \frac{x + 1}{4} = 1

Find Least Common Multiplier of

5, 4 : 20

Multiply by LCM=

20

\frac{x ^{2} + 1}{5} \cdot 20 - \frac{x + 1}{4} \cdot 20 = 1 \cdot 20

Simplify

4 (x^{2} + 1) - 5 (x + 1) = 20

Expand

4 (x^{2} + 1) - 5 (x + 1) : 4 x^{2} - 5 x - 1

4 x^{2} - 5 x - 1 = 20

Move

20

to the left side

4 x^{2} - 5 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 ( - 21 )}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 (- 21) = 19

x_{1, 2} = \frac{- ( - 5 ) \pm 19}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 19}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 19}{2 \cdot 4}

x = \frac{- ( - 5 ) + 19}{2 \cdot 4} : 3

x = \frac{- ( - 5 ) - 19}{2 \cdot 4} : - \frac{7}{4}

The solutions to the quadratic equation are:

x = 3, x = - \frac{7}{4}

4 (x + 10)^{2} + 14 = 50

25 X^{2} - 1 = 0

y, x^{2} + y^{2} - 2 x + 4 y + 5 = 0

4 (x + 2)^{2} = 24

x, 16 x^{2} + 9 = 5

What are the solutions to the equation (x^2+1)/5-(x+1)/4 =1 ?
The solutions to the equation (x^2+1)/5-(x+1)/4 =1 are x=3,x=-7/4
Find the zeros of (x^2+1)/5-(x+1)/4 =1
The zeros of (x^2+1)/5-(x+1)/4 =1 are x=3,x=-7/4