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if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points? I get a relation $a+b+c = 0$ I tried substituting this in given polynomial and tried finding its discriminant but cannot figure out how to find whether it is positive, negative or zero? can someone help?

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    $\begingroup$ Can you show us your work for finding the discriminant? $\endgroup$ Mar 27, 2016 at 12:52

6 Answers 6

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So you have $a+b+c=0$ and the number of intersections between $y$ and the $x$ axis depends on the sign of the discriminant in this case $\Delta=(3b)^2-4(4a)(2c)$. Inserting $b=-a-c$ we get $$\Delta=(3(-a-c))^2-4(4a)(2c)\\ =9(a+c)^2-4(4a)(2c)\\ =9a^2+9c^2-14 a c$$ which attains a minimum value of 0 at $a=0,c=0$ and since you've excluded $a=0$ $\Delta$ is strictly positive so there are always two intersections with the $x$ axis.

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  • $\begingroup$ can we also find whether the roots will be rational or irrational somehow? $\endgroup$
    – Matt
    Mar 27, 2016 at 13:06
  • $\begingroup$ @Raghav That depends on the values of $a$ and $c$, which could be anything. $\endgroup$ Mar 27, 2016 at 13:07
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Since $c=-a-b$, the discriminant $\Delta$ is $$\begin{align}\Delta&=(3b)^2-4\cdot 4a\cdot 2c\\&=9b^2-32a(-a-b)\\&=32a^2+32ab+9b^2\\&=32a^2+32ab+8b^2+b^2\\&=8(4a^2+4ab+b^2)+b^2\\&=8(2a+b)^2+b^2\\&\gt 0\end{align}$$ because $\Delta=0\iff 2a+b=b=0\iff a=b=0$, which contradicts $a\not=0$.

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Case 1: $\; c=0 \implies b=-a$

$4ax^2+3bx+2c \equiv ax(4x-3)$ which has zeroes of $0$ and $\frac{3}{4}$.


Case 2: $\;ax^2+bx+c\equiv a(x-1)^{2} \implies a=c=-\frac{b}{2}$

$4ax^2+3bx+2c \equiv 2a(2x-1)(x-1)$ which has zeroes of $\frac{1}{2}$ and $1$.


Case 3: $\; a\xi^2+b\xi+c=0 \:$ where $\: \xi \neq 0,1$

Considering $f(x)=x^2(ax^2+bx+c)$ with $a+b+c=0$.

$f(0)=f(1)=f(\xi)=0$

By Rolle's Theorem,

$\exists \lambda, \mu \in (\min[0,\xi],\max[1,\xi])$ such that $f'(\lambda)=f'(\mu)=0$ where $\lambda$, $\mu$, $0$, $1$, $\xi$ are all distinct.

$ \implies \left \{ \begin{eqnarray*} 4a\lambda^{3}+3b\lambda^{2}+2c\lambda &=& 0 \\ 4a\mu^{3}+3b\mu^{2}+2c\mu &=& 0 \end{eqnarray*} \right.$

That is $\lambda,\mu$ are the zeroes of $4ax^{2}+3bx+2c$


Hence $\boldsymbol{4ax^{2}+3bx+2c}$ has two distinct zeroes.

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$a = -(b+c)$

The discriminant of the second quadratic is therefore:

$9b^2 - 32ac = 9b^2 + 32bc + 32c^2 = f(b,c)$

where we defined a new function $f(b,c)$

You can easily show that $f(b,c)$ is always positive by considering its own discriminant (which is always negative) and noting that at least one of $b$ or $c$ is nonzero.

Hence the second quadratic has exactly two real roots.

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You found that $a+b+c=0$, which means that $c=-a-b$, so we have $y=4ax^2+3bx-(2a+2b)$. The discriminant is as follows:

$$(3b)^2-4(4a)(-(2a+2b))=9b^2+16a(2a+2b)=9b^2+32a^2+32ab$$

Now, if we plug in $a=1$ and $b=1$ into the discriminant, we clearly get a positive number, meaning there are two solutions. We need to prove that we get a positive number for all $b \in \Bbb{R}$ and all $a \in \Bbb{R}-\{0\}$.

In order to do this, let's try to express the discriminant as the sum of two squares. We can do this by completing the square with $32a^2+32ab$. First, factor out a $32$ to get $32(a^2+ab)$. Then, divide the coefficient of $ab$ by $2$ to get $\frac 1 2$. This will be the coefficient of $b$ in the monomial squared. Then, square that to get $\frac 1 4$ to get the coefficient of $b^2$. Finally, add $\frac 1 4 b^2$ inside the parentheses to get the following:

$$32(a^2+ab+\frac 1 4 b^2)=32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$$

We now know that $32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$. However, the discriminant is $32a^2+32ab+9b^2$ which can be expressed as the sum of two squares as follows:

$$32a^2+32ab+9b^2=32a^2+32ab+8b^2+b^2=32\left(a+\frac 1 2 b\right)^2+b^2$$

Thus, our discriminant is the sum of two squares with one of the squares being multiplied by a positive number, which means that the discriminant is always non-negative. The only case where it is $0$ is if $b=0$ and $a+\frac 1 2 b=0$. Combining these two equations finds that $a=0$, which you said is not under the constraints. Therefore, the discriminant is always positive.

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If $1$ is a root of $ax^2 + bx + c = 0$ Then \begin{align} ax^2 + bx + c &= a(x-1)(x - \lambda) \\ &= ax^2 - a(1 + \lambda)x + a\lambda \\ \end{align}

So $b = -a(1+\lambda)$ and $c = a\lambda$.

The descriminant of $4ax^2 + 3bx + 2c$ is

\begin{align} 9b^2 - 32ac &= 9a^2(1+\lambda)^2 -32a^2\lambda \\ &= a^2(9\lambda^2-14\lambda+9) \\ &= 9a^2\left[\left(x- \dfrac 79 \right)^2 + \dfrac{32}{9}\right] \end{align}

Which is positive.

Hence the polynomial has two real roots.

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