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given that alpha and beta are roots of the quadratic equation ax²+bx+c=0, show that alpha+beta=-6÷a and alphabeta=c÷a​

Sagot :

Answer:

[tex]\alpha + \beta = -\frac{b}{a}[/tex]

[tex]\alpha \beta = \frac{c}{a}[/tex]

Step-by-step explanation:

Given

[tex]ax^2 + bx + c = 0[/tex]

[tex]Roots: \alpha \& \beta[/tex]

Required

Show that:

[tex]\alpha + \beta = -\frac{b}{a}[/tex]

[tex]\alpha \beta = \frac{c}{a}[/tex]

[tex]ax^2 + bx + c = 0[/tex]

Divide through by a

[tex]\frac{a}{a}x^2 + \frac{b}{a}x + \frac{c}{a} = \frac{0}{a}[/tex]

[tex]x^2 + \frac{b}{a}x + \frac{c}{a} = 0[/tex]

The general form of a quadratic equation is:

[tex]x^2 - (Sum)x + (Product) = 0[/tex]

By comparison, we have:

[tex]-(Sum)x = \frac{b}{a}x[/tex]

[tex]-(Sum) = \frac{b}{a}[/tex]

Sum is calculated as:

[tex]Sum = \alpha + \beta[/tex]

So, we have:

[tex]-(\alpha + \beta) = \frac{b}{a}[/tex]

Divide both sides by -1

[tex]\alpha + \beta = -\frac{b}{a}[/tex]

Similarly;

[tex]Product = \frac{c}{a}[/tex]

Product is calculated as:

[tex]Product = \alpha * \beta[/tex]

So, we have:

[tex]\alpha * \beta = \frac{c}{a}[/tex]

[tex]\alpha \beta = \frac{c}{a}[/tex]