If r, s are non zero roots of a0+a1 x+a2 x2=0(a0, a1, a2 ∈ R. and .a2 ≠ 0), then the equality a0+a1 x+a2 x2=a0(1-(x/r))(1-(x/s)) holds

Question

If r, s are non zero roots of a0+a1 x+a2 x2=0(a0, a1, a2 ∈ R. and .a2 ≠ 0), then the equality a0+a1 x+a2 x2=a0(1-(x/r))(1-(x/s)) holds (A) for all values of x, a0 ≠ 0 (B) only when x=0 (C) only when x = r or x = s (D) only when x=r or x=s, a0 ≠ 0

Tardigrade Admin · Accepted Answer

We have r+s=-(a1/a2) ; r s=(a0/a2) Simplifying the R.H.S. of the equation, we have a0(1-(x/r))(1-(x/s)) =a0(1-(x/r)-(x/s)+(1/r s) x2)=a0(1-((r+s/r s)) x+(1/r s) x2) =a0(1+(a1/a0) x+(a2/a0) x2)=a0+a1 x+a2 x2 Thus, the answer is (A) for all values of x, a0 ≠ 0

Q. If $r, s$ are non zero roots of $a_{0} + a_{1} x + a_{2} x^{2} = 0 (a_{0}, a_{1}, a_{2} \in R$ and $a_{2} \neq = 0)$ , then the equality $a_{0} + a_{1} x + a_{2} x^{2} = a_{0} (1 - \frac{x}{r}) (1 - \frac{x}{s})$ holds

for all values of $x, a_{0} \neq = 0$

only when $x = 0$

only when $x = r$ or $x = s$

only when $x = r$ or $x = s, a_{0} \neq = 0$

Q. If r,s are non zero roots of a0​+a1​x+a2​x2=0(a0​,a1​,a2​∈R and a2​=0), then the equality a0​+a1​x+a2​x2=a0​(1−rx​)(1−sx​) holds

for all values of x,a0​=0

only when x=0

only when x=r or x=s

only when x=r or x=s,a0​=0

Q. If $r, s$ are non zero roots of $a_{0} + a_{1} x + a_{2} x^{2} = 0 (a_{0}, a_{1}, a_{2} \in R$ and $a_{2} \neq = 0)$ , then the equality $a_{0} + a_{1} x + a_{2} x^{2} = a_{0} (1 - \frac{x}{r}) (1 - \frac{x}{s})$ holds

for all values of $x, a_{0} \neq = 0$

only when $x = 0$

only when $x = r$ or $x = s$

only when $x = r$ or $x = s, a_{0} \neq = 0$