Question 1036507
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Find the values of m for which the equation {{{x^2 + (m - 2)x+4 = 0}}}
A) has exactly 2 {{{highlight(real)}}} solutions
B) has exactly 1 {{{highlight(real)}}} solution
C) has exactly 0 {{{highlight(real)}}} solutions
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It is about the discriminant of the quadratic equation.
The discriminant in this case is d = b^2 - 4ac,  where

a = 1, b = (m-2), and c = 4.

So, d = (m-2)^2 - 4*4 = {{{m^2 -4m -12}}}.


A)  The equation {{{x^2 + (m - 2)x+4}}} = {{{0}}}  has exactly 2 real solutions if and only if d > 0.

    It is equivalent to an inequality  {{{m^2 -4m -12}}} > {{{0}}}, which has the solutions  m < -2  and/or  m > 6.



B)  The equation {{{x^2 + (m - 2)x+4}}} = {{{0}}}  has exactly 1 real solutions if and only if d = 0.

   It is equivalent to an equation  {{{m^2 -4m -12}}} = {{{0}}},  which has two solutions m = 6  and/or  m = -2.


C)  The equation {{{x^2 + (m - 2)x+4}}} = {{{0}}}  has exactly 0 real solutions if and only if d < 0.

    It is equivalent to an inequality  {{{m^2 -4m -12}}} < {{{0}}}, which has the solutions  -2 < m < 6.


<U>Answer</U>. A) has exactly 2 real solutions  if and only if  m < -2  and/or  m > 6.
        B) has exactly 1 real solution   if and only if  m = -2  and/or  m = 6.
        C) has exactly 0 real solutions  if and only if  -2 < m < 6.
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