Question

The quadratic equation $\frac{\left(x+b\right)\left(x+c\right)}{\left(b-a\right)\left(c-a\right)}+\frac{\left(x+c\right)\left(x+a\right)}{\left(c-b\right)\left(a-b\right)}+\frac{\left(x+a\right)\left(x+b\right)}{\left(a-c\right)\left(b-c\right)}=1$ has

• A. Two real and distinct roots
• B. Two equal roots
• C. None real complex roots
• D. Infinite roots

Answer 1

The correct option is D Infinite roots

$\frac{(x+b)(x+c)}{(b-a)(c-a)}+\frac{(x+c)(x+a)}{(c-b)(a-b)}+\frac{(x+a)(x+b)}{(a-c)(b-c)}=1$

$\Rightarrow -\frac{(x+b)(x+c)}{(a-b)(c-a)}-\frac{(x+c)(x+a)}{(b-c)(a-b)}-\frac{(x-a)(x+b)}{(c-a)(b-c)}=1$

$\Rightarrow \frac{(x+b)(x+c)}{(a-b)(c-a)}+\frac{(x+c)(x+a)}{(b-c)(a-b)}+\frac{(x+a)(x+b)}{(c-a)(b-c)}=-1$

$\Rightarrow (x+b)(x+c)(b-c)+(x+c)(x+a)(c-a)+(x+a)(x+b)(a-b)=-\left[\right(a-b\left)\right(b-c\left)\right(c-a\left)\right]$

$\Rightarrow (x+b)(x+c)(b-c)+(x+c)(x+a)(c-a)+(x+a)(x+b)(a-b)+(a-b)(b-c)(c-a)=0$

$\Rightarrow x^2[b-c+c-a+a-b]+x\left[\right(c+b\left)\right(b-c)+(c+a\left)\right(c-a)+(a+b\left)\right(a-b)+bc(b-c)$
$+ac(c-a)+ab(a-b)+(a-b)(b-c)(b-c)(c-a)=0$

$\Rightarrow x[b^2-c^2+c^2-a^2+a^2-b^2]+bc(b-c)+ac(c-a)+ab(a-b)+(a-b)(b-c)(c-a)=0$

$\Rightarrow b^{2}c-b{c}^2+a{c}^2-a^{2}c+a^{2}b-a{b}^2+abc-a^{2}b-a{c}^2+a^{2}c-b^{2}c+b^{2}a+b{c}^2-bca=$

$\Rightarrow 0=0$

Hence option D is correct, it will have infinite roots.