SOLUTION: Show that the straight line {{{x+y=p}}} will intersect the curve {{{x^2-6x+y^2-8y+4=0}}} at two distinct point if {{{p^2-14p+7<0}}}. {{{I }}} need {{{ STEPS}}}. THANKS!

Question 709188: Show that the straight line will intersect the curve at two distinct point if . need . THANKS!
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
We will start by solving the line's equation for x:
x = p - y
Then we will substitute this expression for x into the other equation:

Next we simplify:

Combining like terms:

Next we will get this equation in standard form of a quadratic. Rearranging the terms so that the is in front, the two y terms are together and the other terms are at the end (Change to addition and then use the Commutative Property):

Group the y terms (Associative Property):

Factor out y:

To make it look more like the "by" of the standard form I'll use the Commutative Property to move the y in back of what we factored:

Group the "other terms" (Associative Property):

We now have the standard form with...
a = 2
b = ((-2p) + (-2))
c =

A quadratic equation in standard form will have two real solutions if its discriminant, , is positive. IOW:

Replacing the a, b and c in this inequality with the expressions we found earlier we get:

Simplifying...

Each term is divisible by 4 (or -4). Since the problem is looking for an expression with a positive we'll divide by -4. (Of course we must remember the rule that says when you divide an inequality by a negative number you must also reverse the inequality at the same time.) Dividing by -4 (and reversing the inequality) we get:

which is equivalent to what the problem asked you to show, i.e.

In summary, we have shown that if then the discriminant (of the quadratic equation we got when we substituted in for x) will be positive. And if the discriminant is positive then there are two real solutions. And these two solutions represent the y coordinates of the points of intersection. So there will be two points of intersection if .
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