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.

\sqrt{x_1}+\sqrt{x_2}

x^2_1-x^2_2

x^2_1+x^2_2

\dfrac{1}{x_1}+\dfrac{1}{x_2}

Collected in the board: Algebraic equation

Steven Zheng posted 2 hours ago

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}

Steven Zheng posted 2 hours ago

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