Question

If r1 and r2 are the roots of ax2 + bx + c = 0, show that r1r2 = ca.If r1 and r2 are roots of ax2 + bx + c = 0, then their values are r1 = −b + b2 − 4ac2a and r2 = −b − b2 − 4ac2ar1r2= + b2 − 4ac2a · −b − b2 − 4ac2a= + b2 − 4ac · −b − b2 − 4ac4a2=2 − b2 − 4ac2 4a2= − (b2 − 4ac)4a2 = 4a2 =ca

Answer 1

Step 1

Given;

$\begin{gathered} r_1r_2=\frac{c}{a} \\ r_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ r_2=\frac{-b-\sqrt{b^2-4ac}}{2a} \end{gathered}$

Step 2

Multiply both terms and fill in the blanks as the steps go on

$\begin{gathered} r_1r_2=(\frac{-b+\sqrt{b^2-4ac}}{2a})\times(\frac{-b-\sqrt{b^2-4ac}}{2a}) \\ First\text{ blank=-b} \end{gathered}$$\begin{gathered} 2\text{a\lparen2a\rparen =4a}^2 \\ Hence, \\ =\frac{(-b+\sqrt{b^2-4ac})\times(-b-\sqrt{b^2-4ac})}{4a^2} \\ Second\text{ blank=-b} \end{gathered}$$\begin{gathered} Expand\text{ the numerator using difference of two squares; \lparen a}^2-b^2)=(a+b)(a-b) \\ a=-b,b=\sqrt{b^2-4ac} \\ (-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac})=(-b)^2-(\sqrt{b^2-4ac})^2 \\ =\frac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2} \\ \end{gathered}$

The third blank will be;

$-b^$$\begin{gathered} =\frac{b^2-(b^2-4ac)}{4a^2} \\ Fourth\text{ blank=b}^2 \end{gathered}$$\begin{gathered} =\frac{4ac}{4a^2} \\ =\frac{c}{a} \\ Last\text{ blank = 4ac} \end{gathered}$