SAJC Check your Understanding Equations teacher

Equations: Check your Understanding Section 1: Solving Quadratic Equation 1. CJC Prelim 8865/2018/Q2 Given that 422 1 0xx   , use the substitution 2ux to find the exact values of x. [4] Answer: 1 2 x Soln: 1 Substituting 2ux into 422 1 0xx   ,    22 1 0 1 2 1 0 uu uu       11 or 2uu    2 2 2 11 reject 0 or 2x x x    2 1 2x  1 2 x 2. TJC Prelim 8865/2018/Q2(a) By means of the substitution ux , and without the use of a graphing calculator, find the value of x which satisfies the equation 856 x. x  [3] Answer: x = 4 Soln: 2    2 885 6 5 6 5 6 8 0 5 4 2 0 4 or 25 4 (Reject) or 25 4 xu ux uu uu uu xx x                

Section 2: Nature of Roots of a Quadratic Equation 1. RI Prelim 8865/2018/Q3(a) Find the set of values of k for which the equation 3 x 2  (k  1) x  3 = 0 has real roots. [4] Answer: {k  I R: k   5 or k ≥ 7} Soln: Since 3 x 2  (k  1) x  3 = 0 has real roots, Discriminant ≥ 0  (k  1) 2  4 (3) (3) ≥ 0  (k  1) 2  (6) 2 ≥ 0  (k  5) (k  7) ≥ 0 Hence the set of values of k is {k  I R: k   5 or k ≥ 7} 2. MJC Prelim 8865/2018/Q2(a) Find the range of values of p for which the equation 2 20x px   has no real roots. [2] Answer: 2 2 2 2p   Soln: 2 Since 2 20x px   has no real roots, Discriminant < 0    2 4(1)(2) 0 8 8 0 88 2 2 2 2 p pp p p           3. PJC Prelim 8865/2018/Q2 Given that p and q are real numbers, show that the equation, 2( )( 2)x p x q   has real roots. State the conditions for the roots to be equal. [3] Answer: p =2 and q = 0 Soln:   2( )( 2) 1x p x q   22 ( 2) 2 0x p x p q     2 2 2 2 2 2Discriminant ( 2) 4(1)(2 ) 4 4 8 4 ( 2) 4 0 since and are real numbers p p q p p p q p q pq              Therefore the roots of the equation are real. For (1) to have equal roots, Discriminant = 0. Therefore, p =2 and q = 0

Section 3: Quadratic expression to be always positive ot always negative 1. SRJC Prelim 8865/2018/Q2(a) Find the exact range of values of k for which   2 13y x k x    is always positive. [3] Answer: 1 2 3 < 1 2 3k   Soln: Since   2 13y x k x    is always positive, Discriminant < 0 and coefficient of x2 = 1 > 0,   2 1 4(1)(3) 0k   Method 1 2 2 11 0kk   Consider 2 2 11 0kk   .   2 2 2 4(1)( 11) 2(1)k     1 2 3 or 1 2 3kk    1 2 3 1 2 3 Hence, 1 2 3 < 1 2 3k   Method 2     22 1 12 0k     1 12 1 12 0kk        1 2 3 1 2 3 0kk         1 2 3 1 2 3 Hence, 1 2 3 < 1 2 3k   2. CJC Prelim 8865/2018/Q1 Find, algebraically, the range of values of k for which 2 40kx x k   for all real values of x. [4] Answer: 2k  Soln: Since 2 40kx x k   ,   20 and coefficient Discr of iminan 1t 0xk           2 2 2 4 4 0 16 4 0 40 2 2 0 kk k k kk          2 or 2 2kk  Combining (1) and (2), 2k  .

3. IJC Prelim 8865/2018/Q1 Find algebraically the set of values of k for which     22 2 2 3 0k x kx k     for all real values of x. [4] Answer:  : , 2  k k k Soln: Since    22 2 2 3 0k x kx k     ,   20 and coefficient of Discriminan 21t 20x k k                2 22 2 2 2 4 2 2 3 0 4 8 4 24 0 4 4 24 0 60 3 2 0 k k k k k k kk kk kk                    2 or 3kk   Combining (1) and (2), 2k  Therefore, the set of solution is  :2kk  4. ACJC Prelim 8865/2018/Q1 Find, algebraically, the set of exact values of m for which 23 24 7 0mx x m   for all real values of x. [4] Answer: 34 7m Soln: 1 Since 23 24 7 0mx x m   ,   20 and coefficient of 3 0 0 1Discriminant x m m       2 2 2 2 ( 24) 4(3 )(7 ) 0 84 > 57 6 84 576 > 0 48 > 0 7 48 48 > 0 77 334 or 4 277 mm m m m mm mm                 3 k

Combining (1) and (2), 34 7m Hence the set of values of m is 3:4 7mm  5. EJC Prelim 8865/2018/Q1 Show that there are no real values of k for which 2( 8) 2 2k x k x   is always negative. [4] Soln:   2 2 2 Discriminant ( 8) 4 2 2 16 64 16 64 > 0 for all real values of kk k k k kk          Therefore discriminant will never be less than 0 and there are not real values of k for 2( 8) 2 2k x k x   is always negative Section 4: Intersection Problems leading to Quadratic Equation 1. JJC Prelim 8865/2018/Q3 Curve C has equation 22( 1) 2xy   . (i) Find the range of values of p if the line y x p does not intersect C. [4] (ii) Deduce the values of p if the line y x p is a tangent to C. [1] Answer: (i) 1 or 3pp  , (ii) 1 or 3pp  Soln: (i) (ii) Sub y x p into 22( 1) 2xy   : 22( 1) ( ) 2x x p    2 2 22 1 2 2x x x px p       222 (2 2 ) ( 1) 0 1x p x p     Since the line does not intersect C, (1) has no real roots. Discriminant 0 22(2 2 ) 4(2)( 1) 0pp    224 4 8 8 8 0p p p     24 8 12 0pp    2 2 3 0pp   ( 3)( 1) 0pp   1 or 3pp  1 or 3pp 

2. MJC Prelim 8865/2018/Q2(b) Find the range of values of k for the line y = 2 x + 3 to intersect the curve   2 2 1 3y kx k x    at least once. [4] Answer: 4.95 or 0.0505kk  Soln:   2 2 1 3 2 3kx k x x           2 2 2 1 3 2 3 0 2 1 6 0 1 kx k x x kx k x           Since the line intersect the curve at least once, (1) has real solutions. Discriminant 0     2 2 2 2 1 4 6 0 4 4 1 24 0 4 20 1 0 kk k k k kk              Using GC, 4.95 or 0.0505 3 s.f.kk  3. DHS Prelim 8865/2018/Q1 Show algebraically that for all real non-zero ,k the line 1y kx intersects the curve 2 3 2 0yx   at two distinct points. [4] Soln: 1 Substitute 1y kx into 2 3 2 0yx   ,     2 22 1 3 2 0 2 3 1 0 kx x k x k x         Discriminant =      2 22 3 4 1kk   2 2 2 2 8 12 9 389 2 3989 4 16 398 0 for all 42 kk kk k kk                Alternative:        2 2 2 2 Discri 2 3 4 1 2 mi 3 4 0 for na all nt kk k k k          Therefore,   22 2 3 1 0k x k x    has 2 distinct real roots for all k . Hence, the line 1y kx will always intersect the curve 2 3 2 0yx   at two distinct points. −4.95 −0.0505 + + −

4. SAJC Prelim 8865/2018/Q2 Two curves are given by the equation 2f ( ) 2 3x ax x   and g( ) 4x ax  respectively where a . Show algebraically that these two curves will intersect each other at 2 distinct points for all real values of a. [3] Soln: 4 At intersection point(s): Equate the 2 equations: 2 2 3 4ax x ax    Rearrange: 2 ( 2) 1 0ax a x    …(1) Discriminant 2 2 ( 2) 4( )(1) 4 4 4 aa a a a        2 4 0 for all real values of aa   The 2 curves will intersect each other at 2 distinct points. 5. VJC Prelim 8865/2018/Q1 The curve   265y k x x   has a minimum point. Find algebraically the set of values of k for which the curve intersects the line 3y