PROBLEM 2. Find the (unique) quadratic polynomial such that all three of the following are true: . All the coefficients

[answerhappygod]

Problem 2 Find The Unique Quadratic Polynomial Such That All Three Of The Following Are True All The Coefficients 1 (34.57 KiB) Viewed 51 times

Problem 2 Find The Unique Quadratic Polynomial Such That All Three Of The Following Are True All The Coefficients 2 (26.15 KiB) Viewed 51 times

Problem 2 Find The Unique Quadratic Polynomial Such That All Three Of The Following Are True All The Coefficients 3 (26.12 KiB) Viewed 51 times

Problem 2 Find The Unique Quadratic Polynomial Such That All Three Of The Following Are True All The Coefficients 4 (30.85 KiB) Viewed 51 times

PROBLEM 2. Find the (unique) quadratic polynomial such that all three of the following are true: . All the coefficients are integers. • The coefficient of x² is 4. 7+ √6 is one of the roots. • Student Solution 1: (x+ (7+√6))(x-(7-√6)) x² + (7²) (7=√6)x= (7+√6) (7-√6) (7 + 86-7-86) X 49+7√5-76456 (43) 4x² - 43 I know •Know that this does not satisfy all the Conditions. We must use some formula/ technique that includes that a = 4 and that the root = 7+√6.
Student Solution 2: X-(7+√6)=0 X =(7+√6) 4x² + bx + c = 0 4 (7+√√5)² + b(7+√√6) +C =0 4(49+ 6) + b(7+√√) +C =0 +b(7+√√6) +C =0 196 + 24 220 +b(7+√√6) + C = 0 6 (7+√6) + C = - 220 7b+ √6²b +C = -220 4 -6=7 6x -7 Student Solution 3: "4x² +bx+c 4x² - 7x + ( 20 I dari know 6²-4ac = 96 6²-161 =96 49-166=96 161 = 45 (=-45 16 ==
Student Solution 4: ? AKS. 4x²+bx+c b must be-7 -bol 4x²-7x+C 7₁ 149-1², so √49-166 = 8√6, 49-16€=384 -16C-335 264 .42 (= -315 38-4 335 4x²-7x-335 46² (2-19 NG) 4/x-(2015)(x-17-√5 ||) == 4/x2 - (2+6)x+12+55) ++) -14x+ bx/ 4(x²-3x +49) == 4x² - 32x +176 = 0 X. 7-5 4(x + (7+5)) (X + (7-6)) 4 (x² +(7-6)x+ (746)* + [49-75-75-4] 4 (x² + [77-√6) x + (7 +√5]x] +49-6) 4 (x² + [(7-6)x +(3+√5]x] + 43 = 4x² + 14x +43 47²-5679172 Student Solution 5: X= 7+56 [12-06 x 2-5 1 Student Solution 6:
Student Solution 7: 3 Quadratic Polynomials bx +c | a₁b₁ and 2 = Z, a = 4₁ 5₁ = 7+56) a, b, √294 = 4x² - 4 (294) 4x² - 1176 so... 4(x²-14x+49)-24:0 4x²-56x +196-24:0 4-56x +172 O (₁x² + bx +² 1 7+ √√6 = 4 (x - इसपे )(x + (204) Student Solution 8: X: 7.16 x- 7 = 16 (x-7)³ = 6 46x-7): 24 4(x-7)²-24=0 q(x) = 4x².56x +172