What are the solutions to the equation x^2+x+(p(x)^2-7p(x))=0 ?

The solutions to the equation x^2+x+(p(x)^2-7p(x))=0 are x=0,x=(7p-1)/(1+p);\quad p\ne-1

Find the zeros of x^2+x+(p(x)^2-7p(x))=0

The zeros of x^2+x+(p(x)^2-7p(x))=0 are x=0,x=(7p-1)/(1+p);\quad p\ne-1

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x^{2} + x + (p (x)^{2} - 7 p (x)) = 0

x = 0, x = \frac{7 p - 1}{1 + p}; p \neq = - 1

Solution steps

x^{2} + x + (p x^{2} - 7 p (x)) = 0

Factor

x^{2} + x + (p x^{2} - 7 p (x)) : x (x + p x - 7 p + 1)

x (x + p x - 7 p + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + p x - 7 p + 1 = 0

Solve

x + p x - 7 p + 1 = 0 : x = \frac{7 p - 1}{1 + p}; p \neq = - 1

The solutions to the quadratic equation are:

x = 0, x = \frac{7 p - 1}{1 + p}; p \neq = - 1

4 x^{2} + 4 y^{2} - 32 x + 8 y - 32 = 0

4 x^{2} + 4 y^{2} + 20 x - 32 y = 5

x^{2} + y^{2} - 20 x + 16 y + 128 = 0

25 - x^{2} - y^{2} \geq 0

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

What are the solutions to the equation x^2+x+(p(x)^2-7p(x))=0 ?
The solutions to the equation x^2+x+(p(x)^2-7p(x))=0 are x=0,x=(7p-1)/(1+p);\quad p\ne-1
Find the zeros of x^2+x+(p(x)^2-7p(x))=0
The zeros of x^2+x+(p(x)^2-7p(x))=0 are x=0,x=(7p-1)/(1+p);\quad p\ne-1