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=(c)/(a) \\ r_1=(-b+√(b^2-4ac))/(2a) \\ r_2=(-b-√(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=((-b+√(b^2-4ac))/(2a))*((-b-√(b^2-4ac))/(2a)) \\ First\text{ blank=-b} \end{gathered}`
`\begin{gathered} 2\text{a\lparen2a\rparen =4a}^2 \\ Hence, \\ =((-b+√(b^2-4ac))*(-b-√(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=√(b^2-4ac) \\ (-b+√(b^2-4ac))(-b-√(b^2-4ac))=(-b)^2-(√(b^2-4ac))^2 \\ =((-b)^2-(√(b^2-4ac))^2)/(4a^2) \\ \end{gathered}`

The third blank will be;

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