Sample Problems. Lecture Notes Equations with Parameters page 1


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1 Lecture Notes Equations with Parameters page Sample Problems. In each of the parametric equations given, nd the value of the parameter m so that the equation has exactly one real solution. a) x + mx m 0 b) mx + mx + m x + c) x + m x. Consider the parametric equation 5m 3x 5mx + 6m + x 0. a) Find all values of m for which x 7 is a solution of the equation. b) Find all values of m for which there are two di erent real solutions of this equation. c) Find all values of the parameter m for which the two solutions of the equation add up to 3. d) Find all values of the parameter m for which the product of the two solutions of the equation is Consider the equation x + mx + 3mx + 5x. a) Find all values of m for which the equation has exactly one real solution. b) Find all values of m for which the equation has no real solution. c) Find all values of m for which the two real solutions x and x are such that x + x :. Consider the parametric equation m + mx + mx x a) Find all values of m for which x is a solution of the equation. b) Find all values of m for which there are two real solutions of the equation. c) Find all values of m for which there are two real solutions, x and x of the equation such that x + x 5. c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
2 Lecture Notes Equations with Parameters page Sample Problems Answers. a) 0; 6 b) 0; c) 0;. a) ; b) m 6 5 c) d) ; a) 5 ; 3; 7 b) (3; 7) c) 3. a) 0 b) ( ; 0) [ 0; c) Sample Problems Solutions. In each of the parametric equations given, nd the value of the parameter m so that the equation has exactly one solution. a) x + mx m 0 0; 6 Solution : Completing the square. x + mx + m {z } x + m x + mx m 0 m m For exactly one solution, we need that m m 0 m 0 m x + m x + mx + m m 0. We solve this equation for m. m 0 multiply by m + 6m 0 m (m + 6) 0 m 0 m 6 Solution : The Quadratic Formula. Based on our equation, we have a, b m, and c m. For exactly one solution, we need the discriminant, b ac to be zero. We solve this equation for m. b ac 0 m () ( m) 0 m + 6m 0 m (m + 6) 0 m 0 m 6 c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
3 Lecture Notes Equations with Parameters page 3 b) mx + mx + m x + 0; Solution : Completing the square. Before we proceed as usual, we have to consider the case for which the equation is linear, namely, m 0. (Remember, we can NOT treat linear equations as a special case of a quadratic equation.) If m 0, our equation is 0 x + which indeed has exactly one solution. Now, if m 6 0; we complete the square mx + x (m ) + m 0 m x + m m x + m 0 m! m x + m (m ) (m ) x + m m m + m 0 m m x + m! (m ) m m + m 0 m To have exactly one solution, the part after the complete square must be equal to zero. (m ) m + m 0 multiply by m m (m ) + m (m ) 0 m + m + m m 0 m 0 m m Solution : The Quadratic Formula. Before we do as usual, we have to consider the case for which the equation is linear, namely, m 0. (Remember, we can NOT treat linear equations as a special case of the quadratic...) If m 0, our equation is 0 x + which indeed has exactly one solution. Now, if m 6 0; we have mx + x (m ) + m 0 and thus a m, b m, and c m. To have exactly one solution, we need the discriminant, b ac to be zero. We solve this equation for m. b ac 0 (m ) (m) (m ) 0 m m + m + m 0 m + 0 m m c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
4 Lecture Notes Equations with Parameters page c) x + m x 0; Solution : Completing the square. First we separately consider the case for which the equation is linear, namely, m 0. If m 0, our equation is x which indeed has exactly one solution. Now, if m 6 0; we multiply both sides by x and complete the square x + m x multiply by x x + m x subtract x x x + m 0 factor out x x + m 0 x x x + 6! x x + {z 6} 6 + m 0! x 6 + m 0 To have exactly one solution, the part after the complete square must be equal to zero. 6 + m 0 multiply by 6 + m 0 m m There is one more possibility we need to consider. The original equation is x+ m x and NOT x x+m 0. We might get exactly one solution as follows: the discriminant is positive, indicating two solutions, but one of the solutions is 0 which would have to be ruled out as it is not in the domain of the original equation. To check out this case, we take x x + m 0 and see what m is so that one of the solutions for x is zero. x x + m 0 (0) 0 + m 0 m 0 We have already considered this case, and so our solution, m 0, m is complete. Solution : Quadratic Formula First we separately consider the case for which the equation is linear, namely, m 0. If m 0, our equation is x which indeed has exactly one solution. Now, if m 6 0; we multiply both sides by x and use the formula: x + m x multiply by x x + m x subtract x x x + m 0 Now a, b, and c m: To have exactly one solution, the discriminant needs to be zero. We solve this equation for m. b ac 0 m ( ) () m 0 m m 0 c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
5 Lecture Notes Equations with Parameters page 5 There is one more possibility we need to consider. The original equation is x + m and NOT x x x + m 0. We might get exactly one solution as follows: the discriminant is positive, indicating two solutions, but one of the solutions is 0 which would have to be ruled out as it is not in the domain of the original equation. To check out this case, we take x x + m 0 and see what m is so that one of the solutions for x is zero. x x + m 0 (0) 0 + m 0 m 0 We have already considered this case, and so our solution, m 0, m is complete.. Consider the parametric equation 5m 3x 5mx + 6m + x 0. a) Find all values of m for which x 7 is a solution of the equation. ; Solution: Just plug in x 7 into the equation and see what that gives us for m. 5m 3x 5mx + 6m + x 0 5m 3 (7) 5m (7) + 6m + (7) 0 5m 35m + 6m m 30m + 0 factor out 6 6 m 5m (m ) (m ) 0 m m b) Find all values of m for which there are two di erent solutions of this equation. m 6 5 Solution: We rearrange the polynomial by degrees of the variable x. x + x ( 5m 3) + 6m + 5m 0. Now a, b 5m 3, and c 6m + 5m. For two solutions, we need the discriminant, b ac to be positive. b ac > 0 ( 5m 3) () 6m + 5m > 0 5m + 30m + 9 m 0m + 6 > 0 m + 0m + 5 > 0 (m + 5) > 0 m 6 5 c) Find all values of the parameter m for which the two solutions of the equation add up to 3. Solution: the sum of the two solutions is b a. b a 3 ( 5m 3) 3 5m m 0 m c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
6 Lecture Notes Equations with Parameters page 6 d) Find all values of the parameter m for which the product of the two solutions of the equation is 0. ; 7 6 Solution: the product of the two solutions is c. Thus we have a c a 0 We solve for m using the quadratic formula: m ; m b p b a ac 5 p m + 5m 0 6m + 5m 0 6m + 5m 0 5 p 5 (6) ( ) (6) 5 p p 5 (6) ( ) m Consider the equation x + mx + 3mx + 5x. 5 a) Find all values of m for which the equation has exactly one real solution. 3 ; 3; 7 Solution: We rearrange the equation and obtain (3m 5) x + (m + ) x + 0. Thus a 3m 5; b m + ; and c. Case. There might be one solution if the equation is linear, i.e. if a 0. This happens when 3m 5 0 3m 5 m 5 3 Then the equation becomes. we write m 5 3 : 3 x + 0 subtract 3 x divide by 3 x 3 Thus m 5 3 gives us one solution. Case. If a 6 0, (when m 6 5 ) then the equation is quadratic. It will have exactly one solution when the 3 discriminant, b ac is zero. This will give us an equation in m. b ac (m + ) () (3m 5) m + m + m + 0 m 0m + (m 3) (m 7) m 3 m 7 c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
7 Lecture Notes Equations with Parameters page 7 Let us check one of these values. If m 3; then the equation is (3 (3) 5) x + (3 + ) x + 0 x + x + 0 (x + ) 0 x and so we have exactly one real solution. Thus there is exactly one solution for x if m 5, 3, or 7. 3 b) Find all values of m for which the equation has no real solution. (3; 7) Solution: there is no solution for a quadratic equation when the discriminant is negative. From the previous part, we have that the discriminant, as a function of m is D (m) (m 3) (m 7) This expression is clearly a quadratic polynomial, with a positive leading coe ecient. regular parabola, which is negative between the two x intercepts. Thus Thus its graph is a (m 3) (m 7) < 0 if and only if 3 < m < 7 y x c) Find all values of m for which the two real solutions x and x are such that x + x : Solution: x + x (x + x ) x x m + 3m 5 3m 5 b c a a (m + ) (3m 5) 3m 5 Thus we need to solve (m + ) (3m 5) 3m 5 for m Recall that in this case m and so 3m (m + ) (3m 5) 3m 5 mulitply by (3m 5) (3m 5) (m + ) (3m 5) subtract 9m 30m + 5 m + m + m + 0 9m 30m + 5 m + m + m + 0 9m 30m + 5 m 6m + 5m m 9 0 (5m 9) (m + ) 0 c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
8 Lecture Notes Equations with Parameters page m 9 m 5 Although we obtained two values for m; they will not both work. Since 3 < 9 5 < 7; the value m 9 5 falls into the interval where there are no real solutions. The only answer is m :. Consider the parametric equation m + mx + mx x a) Find all values of m for which x is a solution of the equation. 0 Solution: We simply substitute x into the equation and solve for m. m + m ( ) + m ( ) ( ) m m + m m 0 b) Find all values of m for which there are two real solutions of the equation. ( ; 0) [ 0; Solution: We have to be careful with parametric equations if the leading term contains a parameter. We always need to treat the linear case separately. Is there a value of m for which the equation is not quadratic? Clearly, if m 0. We check out this case separately. If m 0; then m + mx + mx x 0 + (0) x + (0) x x 0 x x 3 x Thus there is exactly one solution for m 0. Now, if m 6 0, the equation is quadratic. Quadratic equations have two di erent real solutions if the discriminant is positive. 0 mx + x (m ) + m + D b ac (m ) m (m + ) 6 6m D > 0 6 6m > 0 6 > 6m > m Thus it appears, the solution is all values of m that are less than. However, we do need to rule out m 0 since then the equation has only one solution. Thus the answer is m <, m 6 0, or, in interval notation: ( ; 0) [ 0;. c) Find all values of m for which there are two real solutions, x and x of the equation such that x + x 5. Solution: x + x (x + x ) x x b c a a b a (m ) (m + ) (m ) (m + ) (m ) m (m + ) m m m m m m (m ) m (m + ) m m 6m + 6 m m m m 0m + 6 m c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009 c a
9 Lecture Notes Equations with Parameters page 9 m 0m + 6 m 5 m 0m + 6 5m 0 56m + 0m 6 0 7m + 5m 0 (7m ) (m + ) m 7 m Since 7 >, there is no solution for x if m, and thus it is not a solution. On the other hand, falls 7 into the range where the equation has two solutions and so it is correct. We can check: if m, then our equation is Thus a, b 6, and c. Since x + x m + mx + mx x m + ( ) x + ( ) x x b a x x x x x c and a x + x (x + x ) x x 0 x + 6x b c a a 6 ( 6) ( ) For more documents like this, visit our page at and click on Lecture Notes. questions or comments to c copyright Hidegkuti, Powell, 00 Last revised: February 5, 009
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