Question

The equation a"latex-formula">+b$x$+c=0 has roots α, β. Express (α+1)(β+1) in terms of a, b and c.

Answer 1

Answer:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Step-by-step explanation:

We are given the equation:

$ax^2+bx+c=0$

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:

$\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Let x₁ = α and x₂ = β. Substitute:

$\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)$

Combine fractions:

$\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)$

Rewrite:

$\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}$

Multiply and group:

$\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}$

Difference of two squares:

$\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}$

Expand and simplify:

$\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}$

Distribute:

$\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}$

Cancel like terms:

$\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}$

Factor:

$\displaystyle =\frac{4a(a+c-b)}{4a(a)}$

Cancel. Hence:

$\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Therefore:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}$