Question

If a and b are real and $a\ne b$ then show that the roots of the equation, $(a-b)x^2+5(a+b)x-2(a-b)=0$ are real and unequal.

Answer 1

$(a-b)x^2+5(a+b)x-2(a-b)=0$

Here,

$A=a-b$

$B=5(a+b)$

$C=-2(a-b)$

For roots to be real and unequal,

$D>0$

Therefore,

$B^2-4AC$$={\left(5(a+b)\right)}^2-4(a-b)(-2(a-b))$

$=25{(a+b)}^2+8{(a-b)}^2$

$=25(a^2+b^2+2ab)+8(a^2+b^2-2ab)$

$=33a^2+33b^2+34ab>0$

Hence, proved that the given equation have real and unequal roots.