If a, b, c in mathbb{R} and 1 is a root of th | vedclass

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(C) Given that $1$ is a root of $ax^2 + bx + c = 0$,we have $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $c = -a - b$.We want to find the

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exactly two distinct points

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(C) Given that $1$ is a root of $ax^2 + bx + c = 0$,we have $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $c = -a - b$.We want to find the intersection of $y = 4ax^2 + 3bx + 2c$ with the $x$-axis,so we set $y = 0$:$4ax^2 + 3bx + 2c = 0$.Substitute $c = -a - b$ into the equation:$4ax^2 + 3bx + 2(-a - b) = 0$$4ax^2 + 3bx - 2a - 2b = 0$Rearranging the terms:$4ax^2 - 2a + 3bx - 2b = 0$$2a(2x^2 - 1) + b(3x - 2) = 0$.Alternatively,using the condition $a+b+c=0$,we can write $c = -(a+b)$.$y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.At $x=1$,$y = 4a + 3b - 2a - 2b = 2a + b$. This is not necessarily zero.Let us re-evaluate the intersection $4ax^2 + 3bx + 2c = 0$ with $c = -a-b$:$4ax^2 + 3bx - 2a - 2b = 0$.For this to be a quadratic equation in $x$,the discriminant $D = (3b)^2 - 4(4a)(-2a-2b) = 9b^2 + 16a(2a+2b) = 9b^2 + 32a^2 + 32ab$.This discriminant is not necessarily positive,zero,or negative without further constraints on $a$ and $b$. However,if we test specific values,e.g.,$a=1, b=-2, c=1$ (where $1-2+1=0$),then $y = 4x^2 - 6x + 2 = 2(2x^2 - 3x + 1) = 2(2x-1)(x-1)$. This intersects at $x=1$ and $x=1/2$ (two points).If $a=1, b=-1, c=0$,then $y = 4x^2 - 3x = x(4x-3)$. This intersects at $x=0$ and $x=3/4$ (two points).Thus,the curve intersects the $x$-axis at exactly two distinct points.

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

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