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        