Finding value of $m$ which is a part of a quadratic

Question

Q : $$m \gt 2$$ $$x^2 + (m-3)x - 2 = 0$$ If $|x_1 - x_2| = 3$, so $m = ?$

Stuck here. Please give me a hint

Answer 1

Hint Write $$|x_1-x_2|=\sqrt{(x_1+x_2)^2-4x_1x_2}$$

Then you can write $x_1+x_2=-(m-3)$ and $x_1x_2=-2$

Answer 2

You have that the roots of the quadratic polynomial are given by: $$x_1=\frac{-(m-3)^2 + \sqrt{(m-3)^2+8}}{2}$$ and $$x_2=\frac{-(m-3)^2 - \sqrt{(m-3)^2+8}}{2}$$ Therefore, you have that: $$3 = |x_1-x_2| = \sqrt{(m-3)^2+8}$$ You can continue from there.

Answer 3

I would say your condition that the roots are 3 apart you state via the solution formula of the quadratic equation, namely that $$ 3 = \frac{2\sqrt{(m-3)^2-4\cdot1\cdot(-2)}}{2\cdot1} $$ and solve this equation for $m$ under the stated condition on it. Good luck.

Answer 4

Define $f(x,m)=x^2+(m-3)x-2$. When $|x_1-x_2|=3$ we have either $x_1-x_2=3$ or $x_1-x_2=-3$. Start with the first and rewrite we obtain $x_1=x_2+3$. Note that $f(x_2)=0$ and $f(x_1,m)=0$ or $f(x_2+3,m)=f(x_2,m)=0$ which you can write as $f(x_2+3,m)-f(x_2,m)=0$ and simplify to obtain $$3(m+2x_2)=0$$ implying $x_2=-\frac{m}{2}$ which you can plug back in $f(x_2,m)=0$ to obtain $$f(-\frac{m}{2},m)=-\frac14 (m-4)(m-2)=0$$

if you start with $x_1-x_2=-3$ and do the same procedure you obtain exactly the same solutions for $m$.