$\boxed{\text{Problem 1}}$

Find the other solution of the equation $(1+\sqrt3)x^2-(5-\sqrt3)x+6-6\sqrt3=0$ given that $2$ is a solution

Ma solution: $x_1\cdot x_2=\tfrac ca$ therefore letting $x_1=2$ will imply that $x_2=\tfrac c{2a}=3\tfrac{1-\sqrt3}{1+\sqrt3}$

$\boxed{\text{Problem 2}}$

$x^2+3x+c=0$ and $c$ is strictly positive. Find $x_1+x_2$ and $x_1\cdot x_2$ then find the values of $x_1$ and $x_2$ and $c$ given $|x_1|=2$.

So since $x_1$ and $x_2$ are solutions we have $$\left\{\begin{align}x_1+x_2=-b/a\\x_1\cdot x_2=c/a\end{align}\right.$$

Thus: $x_1+x_2=-3$ $(1)$ and $x_1\cdot x_2=c$

Since $|x_1|=2$ then $x_1=2$ or $x_1=-2$, putting this in $(1)$ we get $x_2=-5$ or $x_2=-1$.

Since $c$ is strictly positive then $x_1\neq2$ because then $x_1\cdot x_2$ will be negative.

Putting $x_1=-2$ will give us $x_2=-1$ and will give us $c=2$.

$\boxed{\text{Problem 3}}$

$a,b\in\mathbb{R}$ and $a>0$ and $ax^2+bx-2=0$.

  1. Prove that this equation has two different real solutions without computing them:

$\Delta=b^2-4ac\\= b^2+8a>0$ Since $\Delta>0$ then that equation has two distinct solutions in $\mathbb{R}$.

  1. Prove that the solutions of that equations have different signs without computing them:

Let $x_1$ and $x_2$ be solutions to that equations then we have: $$x_1\cdot x_2=c/a=-2/a$$

We have $-2/a$ is strictly negative since $a>0$ and thus $x_1$ and $x_2$ must have different signs.

  1. Find $b$ given $\frac{2}{x_1}+\frac{2}{x_2}=-7$

With algebraic manipulations we find $x_2+x_1=14/a$ thus b=-14 since $a>0$


In the third question b comes out to be -7.

  • $\begingroup$ proof? $ {}{}{}{}{}$ $\endgroup$ – LoveFood Mar 12 '14 at 9:24
  • $\begingroup$ x1+x2 = 7/a, so b=-7. PS:how to write mathematics here $\endgroup$ – Shashank Mar 12 '14 at 9:27
  • $\begingroup$ $$\frac 2{x_1}+\frac2{x_2}=-7\implies \frac{2(x_1+x_2)}{x_1\cdot x_2}=-7\implies x_1+x_2=(-7/2)\cdot x_1\cdot x_2 \implies x_1+x_2=-7/2 \cdot -2/a$$ So you're right $\endgroup$ – LoveFood Mar 12 '14 at 12:39
  • $\begingroup$ write math using $$ math here $$ $\endgroup$ – LoveFood Mar 12 '14 at 12:39

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