Question

solve it anyone........​

Answer 1

Here $\sf{a,\ b,\ c\neq0.}$

Consider,

$\longrightarrow\sf{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=3}$

$\longrightarrow\sf{\dfrac{a^4bc+b^4ca+c^4ab}{a^2b^2c^2}=3}$

$\longrightarrow\sf{\dfrac{abc(a^3+b^3+c^3)}{(abc)^2}=3}$

$\longrightarrow\sf{\dfrac{a^3+b^3+c^3}{abc}=3}$

$\longrightarrow\sf{a^3+b^3+c^3-3abc=0}$

But,

• $\sf{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}$

Then,

$\longrightarrow\sf{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0}$

If we take $\sf{a+b+c=0,}$ we see,

$\longrightarrow\sf{E_1(1)=a(1)^2+b(1)+c=a+b+c=0}$

$\longrightarrow\sf{E_2(1)=b(1)^2+c(1)+a=b+c+a=0}$

$\longrightarrow\sf{E_3(1)=c(1)^2+b(1)+a=c+b+a=0}$

Hence we get that 1 is the common zero of the quadratic equations $\sf{E_1,\ E_2}$ and $\sf{E_3.}$

If we have to find a quadratic polynomial having zeroes common to $\sf{E_2}$ and $\sf{E_3}$ but different from $\sf{E_1,}$ the other zeroes of

$\longrightarrow\sf{E_2=E_3}$

$\longrightarrow\sf{bx^2+cx+a=cx^2+bx+a}$

$\longrightarrow\sf{bx^2+cx=cx^2+bx}$

$\longrightarrow\sf{bx^2-cx^2=bx-cx}$

$\longrightarrow\sf{x^2(b-c)=x(b-c)}$

$\longrightarrow\sf{x^2(b-c)-x(b-c)=0}$

$\longrightarrow\sf{x(x-1)(b-c)=0}$

Since we're looking for a general condition with coefficients, we take,

$\longrightarrow\sf{b=c}$

Then,

$\longrightarrow\sf{E_2=bx^2+bx+a}$

We know 1 is one zero of $\sf{E_2.}$ So the sum of its zeroes is given by,

$\longrightarrow\sf{1+x'=-\dfrac{b}{b}}$

$\longrightarrow\sf{1+x'=-1}$

$\longrightarrow\sf{x'=-2}$

Hence the other zero of $\sf{E_2}$ and $\sf{E_3}$ each is -2.

Product of zeroes of $\sf{E_2}$ is given by,

$\longrightarrow\sf{\dfrac{a}{b}=-2}$

$\longrightarrow\sf{a=-2b}$

Taking $\sf{x=-2}$ since it's common root,

$\longrightarrow\sf{a=xb}$

$\longrightarrow\sf{a-xb=0}$

Squaring both sides,

$\longrightarrow\sf{(a-xb)^2=0^2}$

$\longrightarrow\sf{a^2-2abx+b^2x^2=0}$

Since $\sf{b=c\implies b^2=bc,}$

$\longrightarrow\sf{a^2-2abx+bcx^2=0}$

Dividing by $\sf{bc}$ since it's non-zero,

$\longrightarrow\sf{x^2-\dfrac{2abx}{bc}+\dfrac{a^2}{bc}=0}$

Taking $\sf{2b=b+c\ \Longleftarrow\ b=c,}$

$\longrightarrow\underline{\underline{\sf{x^2-\dfrac{a(b+c)x}{bc}+\dfrac{a^2}{bc}=0}}}$

Hence (d) is the answer.