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

The solutions to the equation 7x+1=(2x-1)^2 are x= 11/4 ,x=0

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

The zeros of 7x+1=(2x-1)^2 are x= 11/4 ,x=0

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

x = \frac{11}{4}, x = 0

Decimal Notation

x = 2.75, x = 0

Solution steps

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

Expand

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

7 x + 1 = 4 x^{2} - 4 x + 1

Switch sides

4 x^{2} - 4 x + 1 = 7 x + 1

Move

1

to the left side

4 x^{2} - 4 x = 7 x

Move

7 x

to the left side

4 x^{2} - 11 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 \cdot 0 = 11

x_{1, 2} = \frac{- ( - 11 ) \pm 11}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 11}{2 \cdot 4}, x_{2} = \frac{- ( - 11 ) - 11}{2 \cdot 4}

x = \frac{- ( - 11 ) + 11}{2 \cdot 4} : \frac{11}{4}

x = \frac{- ( - 11 ) - 11}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

x = \frac{11}{4}, x = 0

(x - 1)^{2} + 7 = 56

2 b^{2} + 8 b - 19 = 5

- x^{2} + 11 x = 28

18 x - 3 x^{2} = 27

(x - 4)^{3} - (x - 3)^{3} = 343

What are the solutions to the equation 7x+1=(2x-1)^2 ?
The solutions to the equation 7x+1=(2x-1)^2 are x= 11/4 ,x=0
Find the zeros of 7x+1=(2x-1)^2
The zeros of 7x+1=(2x-1)^2 are x= 11/4 ,x=0