If a, b, c in R and 1 is a root of the equati | vedclass

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(C) Given that $1$ is a root of the equation $ax^2 + bx + c = 0$.Therefore,substituting $x = 1$ into the equation,we get $a(1)^2 + b(1) + c = 0$,which

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exactly two distinct points

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(C) Given that $1$ is a root of the equation $ax^2 + bx + c = 0$.Therefore,substituting $x = 1$ into the equation,we get $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $a = -(b + c)$.Now,consider the curve $y = 4ax^2 + 3bx + 2c$.Since $a = -(b + c)$,we substitute this into the expression for $y$:$y = 4(-(b + c))x^2 + 3bx + 2c$$y = -4(b + c)x^2 + 3bx + 2c$However,the problem states $1$ is a root of $ax^2 + bx + c = 0$,so $a + b + c = 0$. This means $c = -(a + b)$.Substituting $c = -(a + b)$ into the curve equation:$y = 4ax^2 + 3bx + 2(-(a + b)) = 4ax^2 + 3bx - 2a - 2b$.Since $a + b + c = 0$,we have $b = -(a + c)$.Actually,a simpler approach: Since $a+b+c=0$,then $c = -a-b$. The equation $ax^2+bx+c=0$ becomes $ax^2+bx-(a+b)=0$.For the curve $y = 4ax^2 + 3bx + 2c$,at the $x-$axis,$y = 0$.$4ax^2 + 3bx + 2c = 0$.Since $a+b+c=0$,$c = -a-b$. Substituting this: $4ax^2 + 3bx - 2a - 2b = 0$.If $x=1$,$4a + 3b - 2a - 2b = 2a + b$. This is not necessarily zero.Let's re-evaluate: $a+b+c=0 \Rightarrow c = -a-b$.$y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.At $x=1$,$y = 4a+3b-2a-2b = 2a+b$. This is not zero unless $2a+b=0$.Wait,the condition $a+b+c=0$ implies $c = -a-b$. The curve is $y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.This is a quadratic equation in $x$. For it to intersect the $x-$axis at exactly one point,the discriminant $D$ must be $0$.$D = (3b)^2 - 4(4a)(-2a-2b) = 9b^2 + 16a(2a+2b) = 9b^2 + 32a^2 + 32ab$.This is not identically zero. Let's re-read: $1$ is a root,so $a+b+c=0$. The curve $y = 4ax^2 + 3bx + 2c$. At $x=1$,$y = 4a+3b+2c = 2a+b+(2a+2b+2c) = 2a+b+2(a+b+c) = 2a+b$. This is not necessarily $0$.Actually,if $a+b+c=0$,then $c = -a-b$. The curve is $y = 4ax^2 + 3bx - 2a - 2b$. This is a parabola. It intersects the $x-$axis where $4ax^2 + 3bx - 2a - 2b = 0$. This is a quadratic equation. It will have roots $x = \frac{-3b \pm \sqrt{9b^2 - 16a(-2a-2b)}}{8a} = \frac{-3b \pm \sqrt{9b^2 + 32a^2 + 32ab}}{8a}$.If $b = -2a$,then $c = -a - (-2a) = a$. The equation $ax^2 - 2ax + a = 0 \Rightarrow a(x-1)^2 = 0$,so $x=1$ is a root. The curve becomes $y = 4ax^2 - 6ax + 2a = 2a(2x^2 - 3x + 1) = 2a(2x-1)(x-1)$. This intersects at $x=1$ and $x=1/2$. Two points.If $a=1, b=0, c=-1$,$y = 4x^2 - 2 = 0 \Rightarrow x^2 = 1/2 \Rightarrow x = \pm 1/\sqrt{2}$. Two points.Thus,the curve intersects at exactly two distinct points.

If $a, b, c \in R$ and $1$ is a root of the equation $ax^2 + bx + c = 0$,then the curve $y = 4ax^2 + 3bx + 2c, a \ne 0$ intersects the $x-$axis at

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