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 $=9=-\frac{b}{a}$, product $=20=\frac{c}{a}$

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: $(x-1)(x-2)=0$

$x=1 \quad or \quad x=2$

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 $1+2+r=6 \implies r=3$

Product $1\cdot2\cdot3=6$, 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 $25-16=9$, $x=\frac{5\pm3}{2}$

$a=4, b=1 \quad (or\ vice\ versa)$

Step 2: $a^2-b^2=(a-b)(a+b)$

Substitute: right side $(a-b)(5)$, left $(a-b)(a+b)= (a-b) \cdot 5$, identity holds.

a=4,\ b=1\ (or\ 1,4)

q8

Step 1: Reciprocal (palindromic) quadratic $ax^2+bx+c$ requires $a=c$.

Here $9x^2 + bx +1$, $9\neq1$, cannot be reciprocal unless scaled differently.

Perhaps roots reciprocal: product of roots $=\frac{1}{9}$, 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\cdot 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 $25+24=49=7^2$

$x=\frac{5\pm7}{2}$

$x=6, \quad x=-1$

Step 2: $P(9)=81 -5\cdot9 -6=81-45-6=30$

zeros: -1,6 \ P(9)=30

q10

Step 1: Zeros of $x^2 -7x + p$ are double (repeated).

Discriminant $=0$: $49 - 4p =0$

$p=\frac{49}{4}$

Root $x=\frac{7}{2}$ (double).

q? Perhaps for the other polynomial $3x^2 -5x +q=0$ or incomplete.

Assuming find p.

p=\dfrac{49{4}}