What are the solutions to the equation a(x)=x^2-4x ?

The solutions to the equation a(x)=x^2-4x are x=0,x=4+a

Find the zeros of a(x)=x^2-4x

The zeros of a(x)=x^2-4x are x=0,x=4+a

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a (x) = x^{2} - 4 x

x = 0, x = 4 + a

Solution steps

a (x) = x^{2} - 4 x

Refine

a x = x^{2} - 4 x

Switch sides

x^{2} - 4 x = a x

Move

a x

to the left side

x^{2} - 4 x - a x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - (4 + a) x = 0

Solve with the quadratic formula

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

Simplify

(- 4 - a)^{2} - 4 \cdot 1 \cdot 0 : - a - 4

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

Separate the solutions

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

x = \frac{- ( - 4 - a ) - a - 4}{2 \cdot 1} : 0

x = \frac{- ( - 4 - a ) - ( - a - 4 )}{2 \cdot 1} : 4 + a

The solutions to the quadratic equation are:

x = 0, x = 4 + a

(x - 3) (8 x + 5) = 0

x^{2} - b^{2} + 4 a^{2} - 4 a x = 0

z^{2} + 2 z - 35 = 0

4 x^{2} + 8 x + 7 = (x + B)^{2}, B

6 x^{2} + 84 x + 284

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