Question

If r, s and t are real numbers and r≠s, then show that the roots of the equation (r- s)x²+7(r + s)x - 3(r-s) 0 are real and unequal.

Answer 1

Answer: Showed.

Step-by-step explanation:  The given quadratic equation is

$(r-s)x^2+7(r+s)x-3(r-s)=0,$

where, 'r' and 's' are real numbers and r ≠ s.

We are to prove that the roots of the equation above are real and unequal. For that, we must show the discriminant greater than 0.

Therefore,

$D=b^2-4ac={7^2(r+s)}^2+12(r-s)^2=49(r^2+2rs+s^2)+12(r^2-2rs+s^2)\\\\\Rightarrow D=61r^2+61s^2+74rs>0, since~~r\neq s.$

Thus, the roots are real and unequal.