What are the solutions to the equation 2(x-3)^2+(x+1)^2=11x+1 ?

The solutions to the equation 2(x-3)^2+(x+1)^2=11x+1 are x=6,x=1

Find the zeros of 2(x-3)^2+(x+1)^2=11x+1

The zeros of 2(x-3)^2+(x+1)^2=11x+1 are x=6,x=1

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2 (x - 3)^{2} + (x + 1)^{2} = 11 x + 1

x = 6, x = 1

Solution steps

2 (x - 3)^{2} + (x + 1)^{2} = 11 x + 1

Expand

2 (x - 3)^{2} + (x + 1)^{2} : 3 x^{2} - 10 x + 19

3 x^{2} - 10 x + 19 = 11 x + 1

Move

1

to the left side

3 x^{2} - 10 x + 18 = 11 x

Move

11 x

to the left side

3 x^{2} - 21 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 3 \cdot 18}{2 \cdot 3}

(- 21)^{2} - 4 \cdot 3 \cdot 18 = 15

x_{1, 2} = \frac{- ( - 21 ) \pm 15}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 15}{2 \cdot 3}, x_{2} = \frac{- ( - 21 ) - 15}{2 \cdot 3}

x = \frac{- ( - 21 ) + 15}{2 \cdot 3} : 6

x = \frac{- ( - 21 ) - 15}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

x = 6, x = 1

\frac{1}{9} - \frac{2}{3} B = \frac{1}{18}

12 x^{2} + 5 x - 2

(x + 5) (x - 6) = 0

\frac{π}{2} - \frac{2 π}{7}

1 < x - \frac{11}{3} + \frac{7}{3}

What are the solutions to the equation 2(x-3)^2+(x+1)^2=11x+1 ?
The solutions to the equation 2(x-3)^2+(x+1)^2=11x+1 are x=6,x=1
Find the zeros of 2(x-3)^2+(x+1)^2=11x+1
The zeros of 2(x-3)^2+(x+1)^2=11x+1 are x=6,x=1