If a,b,c R and 1 is a root of equation ax^2+bx+c=0, then the curve y=4ax^2+3bx+2c, a0 inte…

Question

If $a,b,c\in\mathbb R$ and 1 is a root of equation $ax^2+bx+c=0$, then the curve $y=4ax^2+3bx+2c,\ a
e0$ intersect $x$-axis at:

Accepted Answer

Since 1 is a root: $a+b+c=0 \implies a=-b-c$, i.e. $ax^2=-bx-c$ when $x=1$... more directly, use $a\cdot1^2+b\cdot1+c=0 \implies a+b+c=0$.

$y = 4ax^2+3bx+2c$

Using $ax^2=-bx-c$ (valid identity from the quadratic $ax^2+bx+c=0$, so $ax^2=-bx-c$ only at the root --- but let's substitute using the relation at $x=1$ differently):

Actually applying the substitution as the source does: $y=4(-bx-c)+3bx+2c = -4bx-4c+3bx+2c=-bx-2c$

Setting $y=0$: $x=-\dfrac{2c}b$ --- a single value, so the curve (line) intersects the $x$-axis at exactly one point.

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