3.2 Quadratic equations practice 2

If [math]p=3+\sqrt{2}[/math] and [math]q=4-\sqrt{2}[/math] are the zeros of [math]f(x)=x^2+bx+c[/math] where [math]b[/math] and [math]c[/math] are constants, what is the value of [math]b[/math]?

Consider an equation [math]x^2-bx+12=0[/math] where [math]b[/math] is a positive integer. If the solutions to the equation are integers, how many possible values of [math]b[/math]?

Consider two quadratic equations [math]x^2+4x+3=0[/math] and [math]x^2+bx+12=0[/math]. If these equations have some common solutions, find the maximum possible value of [math]b[/math].

Which of the following equations has/have two positive solutions?

Consider an equation [math]9x^2+24x+a=0[/math] where [math]a[/math] is a positive integer. If the equation has no radical solutions (i.e. [math]b^2-4ac[/math] is a rational number), how many possible values of [math]a[/math]?

If [math]p[/math] and [math]q[/math] are the solutions to the equation [math]2x^2+3x+5=0[/math], what is the value of [math](p-q)^2[/math]?

Mr. Fang has two light bulbs. The first bulb has resistance [math]R_1[/math] ohms and the other one has resistance [math]R_2[/math] ohms. [math]R_1[/math] and [math]R_2[/math] are the solutions of the quadratic equation [math]x^2-5x+1=0[/math]. Mr. Fang connects them in parallel circuit. Let [math]R[/math] be the total resistance. Given the formula of resistance of parallel circuits
$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$$
find the total resistance [math]R[/math].

Let [math]\alpha[/math] and [math]\beta[/math] be the roots of the equation [math]2x^2+6x-5=0[/math]. If the quadratic equation [math]ax-bx-c=0[/math] has solutions [math]\frac{1}{\alpha}[/math] and [math]\frac{1}{\beta}[/math] where [math]a, b[/math], and [math]c[/math] are positive integers, what is the minimum of [math]a[/math]?

If [math]x=ai[/math] is a solution to the equation [math]8ix^2+5i=18x[/math] where [math]a[/math] is a negative real number, what is the value of [math]a[/math]?

Find the negative solution of the equation [math]| \dfrac{3+2xi}{4-3i} | =1[/math]