What are the solutions to the equation x^2-bx+b^2=0 ?

The solutions to the equation x^2-bx+b^2=0 are x= b/2+i(sqrt(3)b)/2 ,x= b/2-i(sqrt(3)b)/2

Find the zeros of x^2-bx+b^2=0

The zeros of x^2-bx+b^2=0 are x= b/2+i(sqrt(3)b)/2 ,x= b/2-i(sqrt(3)b)/2

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x^{2} - b x + b^{2} = 0

x = \frac{b}{2} + i \frac{3 b}{2}, x = \frac{b}{2} - i \frac{3 b}{2}

Solution steps

x^{2} - b x + b^{2} = 0

Solve with the quadratic formula

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

Simplify

(- b)^{2} - 4 \cdot 1 \cdot b^{2} : 3 ib

x_{1, 2} = \frac{- ( - b ) \pm 3 ib}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - b ) + 3 ib}{2 \cdot 1}, x_{2} = \frac{- ( - b ) - 3 ib}{2 \cdot 1}

x = \frac{- ( - b ) + 3 ib}{2 \cdot 1} : \frac{b}{2} + i \frac{3 b}{2}

x = \frac{- ( - b ) - 3 ib}{2 \cdot 1} : \frac{b}{2} - i \frac{3 b}{2}

The solutions to the quadratic equation are:

x = \frac{b}{2} + i \frac{3 b}{2}, x = \frac{b}{2} - i \frac{3 b}{2}

r^{2} = 144

j (x) = - x^{2} + 3 x + 10, j (3 x - 2)

8 = 2 x^{2} - 5

5 x (x - 3) + 9 = - 2 x (x - 3) + 7 x - 2

(x + 9)^{2} = - x - 7

What are the solutions to the equation x^2-bx+b^2=0 ?
The solutions to the equation x^2-bx+b^2=0 are x= b/2+i(sqrt(3)b)/2 ,x= b/2-i(sqrt(3)b)/2
Find the zeros of x^2-bx+b^2=0
The zeros of x^2-bx+b^2=0 are x= b/2+i(sqrt(3)b)/2 ,x= b/2-i(sqrt(3)b)/2