If $$x_1$$ and $$x_2$$ are the two roots of the quadratic equation $$x^2-px+q=0$$, $$(x_1x_2)$$. Find the values of the following expressions.

Question

If $$x_1$$ and $$x_2$$ are the two roots of the quadratic equation $$x^2-px+q=0$$, $$(x_1>x_2)$$. Find the values of the following expressions.

Uniteasy Admin · Accepted Answer

Since $$x_1$$ and $$x_2$$ are two roots of the quadratic equation $$x^2-px+q=0$$, the following relationships are established by using Vieta's formula,
$$x_1+x_2=p$$
and
$$x_1x_2 = q$$
The idea is to transform the following expressions to the form in terms of $$x_1+x_2$$ and $$x_1x_2$$. And then substitute the values.$$\sqrt{x_1}+\sqrt{x_2}  $$
$$=\sqrt{(\sqrt{x_1}+\sqrt{x_2}  )^2} $$
$$=\sqrt{x_1+x_2+2\sqrt{x_1x_2} } $$
$$=\sqrt{p+2\sqrt{q} } $$ $$x_1-x_2$$
$$=\sqrt{(x_1-x_2)^2} $$
$$=\sqrt{(x_1+x_2)^2-4x_1x_2} $$
$$=\sqrt{p^2-4q} $$
Apply the difference of squares identity
$$x^2_1-x^2_2$$
$$=(x_1+x_2)(x_1-x_2)$$
$$=p\sqrt{p^2-4q}$$ $$x^2_1+x^2_2$$
$$=(x_1+x_2)^2-2x_1x_2$$
$$=p^2-2q$$ $$\dfrac{1}{x_1}+\dfrac{1}{x_2}  $$
$$=\dfrac{x_1+x_2}{x_1x_2} $$
$$=\dfrac{p}{q} $$

Question

Answer