Alberta Free Tutoring And Homework Help For Math 20-1


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the Equation  $\left|x^2-px+5\right|=7$|x2px+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

4 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 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

4 years ago

Answered By Xuezhong J

equation ax2+bx+c=0 has real solutions when y= b-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.