Question

$x+y+z=15$ when $a,x,y,z.b$ are in A.P. and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$ when $a,x,y,z,b$ are in H.P., then the quadratic equation whose roots are $\frac{1}{a}$ and $\frac{1}{b}$ is

• A. $x^2-10x+9=0$
• B. $x^2+10x-9=0$
• C. $39x^2+10x-1=0$
• D. $9x^2+10x+1=0$

Answer 1

The correct option is C $39x^2+10x-1=0$

$x+y+z=152y=(x+z)3y=15y=5x+z=10\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}y=\frac{2xz}{(x+z)}\frac{1}{x}+\frac{1}{z}+\frac{1}{2x}+\frac{1}{2z}=\frac{5}{3}\frac{3}{2x}+\frac{3}{2z}=\frac{5}{3}\frac{1}{x}+\frac{1}{z}=\frac{10}{9}\frac{x+z}{xz}=\frac{10}{9}xz=9{(x-z)}^2={(x+z)}^2-4xz{(x-z)}^2={\left(10\right)}^2-4\times 9{(x-z)}^2=100-36{(x-z)}^2=64x-z=\pm 8AP:13,9,5,1,-3AP:-3,1,5,9,13$

if $x-z=8x-(x+2d)=8-2d=8d=-4$

if $x-z=-8x-(x+2d)=-8-2d=-8d=4$

$b=a+4db=a-16b=-3,13y+z=10a+d+a+3d=102a=(10-4d)a=5-2da=13,-3x^2-\frac{a+b}{ab}x+\frac{1}{a}\times \frac{1}{b}=0x^2-\frac{10x}{-39}-\frac{1}{39}=039x^2+10x-1=0$