if the equation below is to have 1 double real root, determine the value for m.

Question

 if the equation below is to have 1 double real root, determine the value for m.

 

(m-2)x^2+mx+2=0

Answer 1

What Damazo said except why did he substitute in 4?

m^2 - 4(m - 2)(2) = 0   If the discriminant is 0, you have one, repeated root. just solve this and you get m=4.

m^2  - 8m + 16 =0

(m - 4)(m - 4) =0

m = 4

Answer 2

Hello, Nick

 If you recall the quadratic formula, the part that is underneath the radical sign ( the sq. rt. sign) is called the Determinant.  The he formula for the Determinant is: 

 

b^2-4ac.   The variables a,b,c come from the quadratic formula when it is written in the standard form, that is

 

ax^2+bx+c= 0.  So, for example if I give you 3x^2+2x-4=0. The a value would be 3, the b value would be 2, and the c value would be -4.

 

The determinant is very important because it tells us whether we have one solution, two solution, or no real solution.  

 

Now, if the determinant= 0, there is only one solution.

         if the determinant is greater than zero, there are two solutions.

         if the determinant is less than zero,there are no real solutions.

 

in the problem that you posted

(m-2)x^2+ mx+ 2= 0

a= (m-2)

b= m

c= 2

 

Now, you want to find a value of m that will yield one solution.  This means that b^2-4ac must equal zero. After some minutes of trial and error, I found of the answer. m must be equal to 4.

 

b^2-4ac                   Substituting using m's

m^2-4(m-2)(2)         Substituting 4 for m

(4)^2-4(4-2)(2)        Evaluating using the order of operations

 16-(4)(2)(2)

   16-16= 0.          

 

So, for this parabola to have one real solution, m must be equal to 4.

 

i hope this helps out. I really enjoyed solving it.  Well keep posting your question. 

D. Y. T.

 

 

Answer 3

Nick,

Use the quadratic formula:

If equation is ax² + bx + c = 0, then the solutions are x = (-b +- square root (b² - 4ac))/2a, where in your problem:

a = (m-2), b = m  & c = 2

 

The discriminant inside the square root is (b²-4ac) or m² - 4(m-2)(2) = m² -8m + 16 = (m - 4)² 

So its square root is just (m-4)

Thus, the two solutions for x in the quadratic formula become:  x = (-m +- (m-4))/2(m-2).  Call them x₁ & x₂

Thus, x1 takes the positive value of the square root, x₁ = (-m + (m-4))/2(m-2) = -4/2(m-2) = -2/(m-2)

And x₂ takes the negative value of the square root , x₂ = (-m -(m-4))/2(m-2)  = (-2m + 4)/2(m-2) = -2(m-2)/2(m-2)

So x₂ = -1 no matter the value of m, but x₁ does depend on the value of m

 

If m = 0, then x₁ = +1 and x₂ = -1 and your equation simplifies to -2x² + 2 =0 or x2 = -2/-2 = +1.

M cannot be +2 because (m-2) in the denominator for x₁ would be zero and x₁ would be infinity

For any other positive or negative integer m there is a two root integer solution for your equation.