Q1. Find a quadratic polynomial whose zeroes are (5-3√2) and (5+3√2).

q1(a) Step 1: Form of quadratic with zeros 3 and -2: P(x)=(x-3)(x+2) Step 2: Expand: (x-3)(x+2)=x^2+2x-3x-6=x^2-x-6 x^2-x-6 q1(b) Step 1: Leading coefficient a=3, zero at -2, and from context zeros 3 and -2: P(x)=3(x-3)(x+2) Step 2: First expand (x-3)(x+2)=x^2-x-6 Step 3: Multiply by 3: P(x)=3(x^2-x-6)=3x^2-3x-18 Verify P(-2)=3(4)-3(-2)-18=12+6-18=0 3x^2-3x-18 q1(c) Step 1: Zeros 5, 4 so sum =9=-(b)/(a), product =20=(c)/(a) Step 2: b=-9a, c=20a P(x)=ax^2-9ax+20a=a(x^2-9x+20) (Or monic: x^2-9x+20) a(x^2-9x+20) q4 (or q1(d)?) Step 1: P(x)=x^2+4x+k=0, zeros m,n By Vieta's formulas: Sum m+n=-4, product mn=k m+n=-4 \\ mn=k q5 Step 1: Verify x=1 is root of x^2-3x+2=0: P(1)=1-3+2=0 Yes. Step 2: Solve x^2-3x+2=0: Factor: (x-1)(x-2)=0 x=1 or x=2 x=1,2 q6 Step 1: Verify zeros 1, 2 for P(x)=x^3-6x^2+11x-6 P(1)=1-6+11-6=0 P(2)=8-24+22-6=0 Yes. Step 2: Third zero: divide or Vieta. Sum zeros 1+2+r=6 r=3 Product 1·2·3=6, matches constant -(-6)=6 (for monic, sum=6, etc.) Factors: (x-1)(x-2)(x-3)=x^3-6x^2+11x-6 Vieta relations verified. zeros: 1,2,3 q7 Step 1: Zeros a,b of r(x)=x^2-5x+4=0 Sum a+b=5, product ab=4 Solve: discriminant 25-16=9, x=(5±3)/(2) a=4, b=1 (or\ vice\ versa) Step 2: a^2-b^2=(a-b)(a+b) Substitute: right side (a-b)(5), left (a-b)(a+b)= (a-b) · 5, identity holds. a=4,\ b=1\ (or\ 1,4) q8 Step 1: Reciprocal (palindromic) quadratic ax^2+bx+c requires a=c. Here 9x^2 + bx +1, 9≠1, cannot be reciprocal unless scaled differently. Perhaps roots reciprocal: product of roots =(1)/(9), but for reciprocal roots product=1, impossible. Perhaps find reciprocal of 9x^2 + bx +1: x^2 P(1/x)=9 + b x + x^2 So reciprocal polynomial is x^2 + b x +9 "of the other" perhaps means it equals another or something. Assuming find b for symmetry, impossible. Perhaps misread, skip specific or assume b such that symmetric after scale. Perhaps problem: the polynomial 9x^2 + b x +1 is reciprocal type if b chosen? No. Perhaps "reciprocal equation" substitute y=x+1/x. Assume divide by x, but since unclear, perhaps b arbitrary or find for roots r,1/r. For roots r,1/r, P(x)=(x-r)(x-1/r)=x^2 -(r+1/r)x +1 To have leading 9: 9(x^2 - s x +1/9) where s=r+1/r No: product roots=1/9 for reciprocal? No, for roots r,1/r product=1. Constant/leading = product =1/9 ≠1, so no such real b for reciprocal roots. Perhaps the problem is to find b so that it is reciprocal polynomial, but impossible. Perhaps "reciprocal polynomial" means the reverse coefficients. The "other" perhaps another polynomial. Perhaps it's "9x^2 + bx +1 is the reciprocal of ax^2 + cx +9" or something. Unclear, perhaps b= something. Looking at text "is reciprocal polynomial of the other." Perhaps "the reciprocal polynomial", find its form. Reciprocal of P(x)=9x^2 + b x +1 is x^2 P(1/x)= 1· x^2 + b x +9 = x^2 + b x +9 x^2 + b x + 9 q9 Step 1: Zeros of P(x)=x^2 -5x -6=0 Discriminant 25+24=49=7^2 x=(5±7)/(2) x=6, x=-1 Step 2: P(9)=81 -5·9 -6=81-45-6=30 zeros: -1,6 \\ P(9)=30 q10 Step 1: Zeros of x^2 -7x + p are double (repeated). Discriminant =0: 49 - 4p =0 p=(49)/(4) Root x=(7)/(2) (double). q? Perhaps for the other polynomial 3x^2 -5x +q=0 or incomplete. Assuming find p. p=(49)/(4)