Answer:
[tex]\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)[/tex]
Step-by-step explanation:
We are given the equation:
[tex]ax^2+bx+c=0[/tex]
Which has roots α and β.
And we want to express (α + 1)(β + 1) in terms of a, b, and c.
From the quadratic formula, we know that the two solutions to our equation are:
[tex]\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]
Let x₁ = α and x₂ = β. Substitute:
[tex]\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)[/tex]
Combine fractions:
[tex]\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)[/tex]
Rewrite:
[tex]\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}[/tex]
Multiply and group:
[tex]\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}[/tex]
Difference of two squares:
[tex]\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}[/tex]
Expand and simplify:
[tex]\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}[/tex]
Distribute:
[tex]\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}[/tex]
Cancel like terms:
[tex]\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}[/tex]
Factor:
[tex]\displaystyle =\frac{4a(a+c-b)}{4a(a)}[/tex]
Cancel. Hence:
[tex]\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)[/tex]
Therefore:
[tex]\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}[/tex]
