Hi there,
just wondering if anyone could please shed some light on this one for me?. If we consider the equation: ax^2+bx+a=0, then:
Prove that IF this equation has only one rational root, then b=+-2a.
Your time is much appreciated!. Thank you kindly..
+118723 just wondering if anyone could please shed some light on this one for me?. If we consider the equation: ax^2+bx+a=0, then:
Prove that IF this equation has only one rational root, then b=+-2a.
OR if you allowed to take short cuts you could say that the quadratic formula gives the roots as
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)
but in your quadratic c=a so
\(x = {-b \pm \sqrt{b^2-4a^2} \over 2a}\)
This will have exactly one solution when
\(b^2-4a^2=0\)
Continue the same as for my first 'proper' solution.
+118723 Hi, 
ax^2+bx+a=0
\(ax^2+bx+a=0\\ ax^2+bx=-a\\ x^2+\frac{b}{a}x = -1\\ x^2+\frac{b}{a}x + (\frac{b}{2a} )^2 = -1 + (\frac{b}{2a} )^2 \\ (x+ \frac{b}{2a} )^2 = -1 + (\frac{b}{2a} )^2 \\ (x+ \frac{b}{2a} )^2 = -\frac{4aa}{4aa} + \frac{bb}{4aa} \\ (x+ \frac{b}{2a} )^2 = \frac{bb-4aa}{4aa} \\ \)
\((x+ \frac{b}{2a} )^2 = \frac{bb-4aa}{4aa} \\ x+ \frac{b}{2a} = \pm \sqrt{ \frac{bb-4aa}{4aa} } \\ x = - \frac{b}{2a} \pm \sqrt{ \frac{bb-4aa}{4aa} } \\ x = - \frac{b}{2a} \pm \frac{1}{2a} \sqrt{ bb-4aa } \\ a\ne0\\ \mbox{There will be only one root when }\\ b^2-4a^2=0 \\ (b-2a)(b+2a)=0\\ b=\pm2a \)
+118723 just wondering if anyone could please shed some light on this one for me?. If we consider the equation: ax^2+bx+a=0, then:
Prove that IF this equation has only one rational root, then b=+-2a.
OR if you allowed to take short cuts you could say that the quadratic formula gives the roots as
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)
but in your quadratic c=a so
\(x = {-b \pm \sqrt{b^2-4a^2} \over 2a}\)
This will have exactly one solution when
\(b^2-4a^2=0\)
Continue the same as for my first 'proper' solution.
