the Equation $\left|x^2-px+5\right|=7$|x2−px+5|=7 has real solutions
A. only when p2 > or equal to 48
B. Only when p2 < 48
C. for any p-values
D. for no values of p
2 years ago
The absolute value sign means the left hand side of the equation is either positive 7 or negative seven. For the first possibility x2-px+5=7 or x2-px=2 or x2=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 x2. For any value of p x will always need to be slightly larger in absolute value.
Similarily for the second case x2-px+5=-7 or x2-px=-12 or x2=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. x2=px-12 becomes x2=-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
equation ax2+bx+c=0 has real solutions when y= b2 -4ac > =0.
the equation in this question equal 2 equations: 1. x2 -px+5=7, 2. x2 -px+5=-7,
for 1, y=(-p)2 -4x1x(-2)>=0, p can be all real numbers.
for 2, y=(-p)2 - 4x1x12>=0, p2 >=48
so consider 1and 2, C is correct, that is p can be any values.