SOLUTION: (1) Find the range of m such that the equation |x^2-3x+2|=mx has 4 distinct real solutions a,b,c,d (2) Express the value of s(m) = 1/a^2 + 1/b^2 + 1/c^2 + 1/d^2 (3) when m varies

Question 880222: (1) Find the range of m such that the equation |x^2-3x+2|=mx has 4 distinct real solutions a,b,c,d
(2) Express the value of s(m) = 1/a^2 + 1/b^2 + 1/c^2 + 1/d^2
(3) when m varies as in (1) find the range of s(m)
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Given;
(1) y = |x^2-3x+2| and
(2) y = mx
The parabola opens upward with symmetry around
(3) x = -b/(2*a) or
(4) x = 3/2
At x = 3/2, (1) gives
(3) y = |1.5^@ -3*1.5 +2| or
(4) y = |2.25 - 4.5 +2| or
(5) y = |-.25| or
(6) y = .25
In order to have 4 distinct crossings (solutions) of (1) and (2) m must be greater than zero and less than
(7) .25 = m*1.5 or
(8) m = 1/6 giving the range of m as
(9) 0 < m < 1/6
When m = 0 we get
(10) {a,b,c,d} = {1,1,2,2} the point on the x axis where the parabola of (1) crosses.
To get the set for m = 1/6 set (1) = (2) at m = 1/6
(11) x^2-3x+2 = x/6
Use the quadratic equation to find
(12) x = {0.0326,0.8840)
These are the values of a and d when m=1/6
The values of b and c are 1/4 when m = 1/6
put tyhe sets into the expression for s(m) and get
(13) s(0) = 2.5 and
(14) s(1/6) = 939.507
Sorry, I made a big error in the above solution. It's a bit harder than I gave you above. Here's the right answer.
My error is in the selection of m = 1/6. The value of m can be slightly greater than 1/6. It turns out to be approximately 0.1715 vs 0.1666.
The straight line mx can increase until it just touches the (inverted) parabola,
(15) y = -x^2+3x-2 between 1 To get m we set the derivitive of (15) equal to the slope of mx and get
(16) -2x + 3 = m or
(17) x = (3-m)/2
Now substitute (17) into (15) to get (after some algebra)
(18) y = (1-m^2)/4
Note that when m=0, (18) gives us y = 1/4, the maximum value of y where the slope is zero.
Now to get the value of m we need to set (18) equal to mx, where x is given by (17),
(19) 1-m^2)/4 = m*(3-m)/2
Simplify (19) to get
(20) m^2 - 6m + 1 = 0
Use the quadratic equation to find
(21) m = 3 +- sqrt(2)
Chose the negative to get
(22) m = 3 - sqrt(2) or
(23) m = 0.1715 which slightly greater than 1/6.
Now that we have m we can the range of m as
(24) 0 We can get b and c, equal at tangent point
(25) (x,y) = ((3-m)/2,(1-m^2)/4) or
(26) b = c = (3-m)/2 or
(27) b = c = (3-(3 - sqrt(2))/2 or
(28) b = c = sqrt(2)/2 or
(29) b = c = 1/sqrt(2)
Now to get a and d we need to set the parabola of (1), without the absolute value sign, because we are outside 1 (30) x^2-3x+2 = (3-sqrt(2))*x, and solve for x using the quadratic equation to get
(31) x = (3-sqrt(2) +- sqrt(9-6*sqrt(2)) or
(32) x = {0.868341,2.3032}
The value in (32) correspond to a and d respective.
Using the values of {a,b,c,d} at m = 0.1715 we get
(33) s(0.1716) = 35.485

RELATED QUESTIONS
Let K be a real number, and consider the quadratic equation (k+1)x^2+4kx+2=0 a. Show... (answered by robertb)

The equation x^2 + k = 6 x has two distinct real roots. Find the range of values of k. (answered by stanbon)

Given these two equations : f(x)=x^2-6x+9 and g(x)=-3x-5, change the slope of the line so (answered by ewatrrr)

Find the values of b such that the equation below always gives real solutions. -3x^2 +... (answered by ikleyn)

If a,b,c,d, and e are distinct prime numbers such that: b=(a-1)^1/2 +1 ; c=(b-1)^1/2 +1 ; (answered by solver91311)

If a,b,c,d, and e are distinct prime numbers such that: b=(a-1)^1/2 +1 ; c=(b-1)^1/2 +1 ; (answered by MathLover1,ikleyn,math_tutor2020)

Find the value of K such that 2x^2-3x+k has each solution set. A. Two real number... (answered by jim_thompson5910)

"The equation x^2-3x+k^2=4 has two distinct real roots. Find the possible values of k."... (answered by josgarithmetic)

I couldn't work letter C and #2, please assist. Please check the rest of the work and... (answered by yuendatlo,solver91311)