the Equation $\left|x^2-px+5\right|=7$|x^{2}−px+5|=7 has real solutions

A. only when p^{2 }> or equal to 48

B. Only when p^{2} < 48

C. for any p-values

D. for no values of p

2 years ago

Answered By Albert S

The absolute value sign means the left hand side of the equation is either positive 7 or negative seven. For the first possibility x^{2}-px+5=7 or x^{2}-px=2 or x^{2}=px+2 as p increase in the positive or negative sense x must have the same sign. If p is negative the domain is x<=0. If p is positive 0>=x. x will always be slightly larger than p because the product px always needs to have 2 added on to be the same value a x^{2}. For any value of p x will always need to be slightly larger in absolute value.

Similarily for the second case x^{2}-px+5=-7 or x^{2}-px=-12 or x^{2}=px-12. As the various options for p are considered it will again need to always have the same sign as the value of x (if p is negative then x must be negative to always give a positive product - (neg)x(neg)=(pos)). x will always be slightly smaller (in absolute value) than p because the product px needs to have 12 subtracted from it to be the same as the the square of x.

The only limitation on p to give real solutions is p cannot equal zero in the second case. x^{2}=px-12 becomes x^{2}=-12 if p =0 and then x must be an imaginary number. If p is negative then x<0 if p is positive x>0

2 years ago

Answered By Xuezhong J

equation ax^{2}+bx+c=0 has real solutions when y= b^{2 }-4ac > =0.

the equation in this question equal 2 equations: 1. x^{2} -px+5=7, 2. x^{2} -px+5=-7,

for 1, y=(-p)^{2} -4x1x(-2)>=0, p can be all real numbers.

for 2, y=(-p)^{2} - 4x1x12>=0, p^{2} >=48

so consider 1and 2, C is correct, that is p can be any values.

2 years ago

## Answered By Albert S

The absolute value sign means the left hand side of the equation is either positive 7 or negative seven. For the first possibility x

^{2}-px+5=7 or x^{2}-px=2 or x^{2}=px+2 as p increase in the positive or negative sense x must have the same sign. If p is negative the domain is x<=0. If p is positive 0>=x. x will always be slightly larger than p because the product px always needs to have 2 added on to be the same value a x^{2}. For any value of p x will always need to be slightly larger in absolute value.Similarily for the second case x

^{2}-px+5=-7 or x^{2}-px=-12 or x^{2}=px-12. As the various options for p are considered it will again need to always have the same sign as the value of x (if p is negative then x must be negative to always give a positive product - (neg)x(neg)=(pos)). x will always be slightly smaller (in absolute value) than p because the product px needs to have 12 subtracted from it to be the same as the the square of x.The only limitation on p to give real solutions is p cannot equal zero in the second case. x

^{2}=px-12 becomes x^{2}=-12 if p =0 and then x must be an imaginary number. If p is negative then x<0 if p is positive x>02 years ago

## Answered By Xuezhong J

equation ax

^{2}+bx+c=0 has real solutions when y= b^{2 }-4ac > =0.the equation in this question equal 2 equations: 1. x

^{2}-px+5=7, 2. x^{2}-px+5=-7,for 1, y=(-p)

^{2}-4x1x(-2)>=0, p can be all real numbers.for 2, y=(-p)

^{2}- 4x1x12>=0, p^{2}>=48so consider 1and 2, C is correct, that is p can be any values.