Algebraic Identities1. \((x+y)^2=x^2+y^2+2xy\)

2. \((x-y)^2=x^2+y^2-2xy\)

3. \(x^2-y^2=(x+y)(x-y)\)

4. \((x+y)^2-(x-y)^2=4xy\)

5. \(x^3+y^3=(x+y)(x^2+y^2-xy)\)

6. \(x^3-y^3=(x-y)(x^2+y^2+xy)\)

QuadraticsThe general form of a quadratic equation is \(ax^2+bx+c=0\). It's roots are:

\(x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}\) and \(x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}\)

Expression \(b^2-4ac\) is called

discriminant:

- If discriminant is positive quadratics has two roots;
- If discriminant is negative quadratics has no root;
- If discriminant is zero quadratics has one root.

When graphed quadratic expression (\(ax^2+bx+c=0\)) gives parabola:

- The larger the absolute value of \(a\), the steeper (or thinner) the parabola is, since the value of y is increased more quickly.

- If \(a\) is positive, the parabola opens upward, if negative, the parabola opens downward.

Viete's theoremViete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).Common mistake to avoidNever reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.For example, \(xy=y\) cannot be reduced by \(y\) because \(y\) could be 0 and we cannot divide by 0. If we do we'll loose one of the solutions. The correct way is: \(xy=y\) --> \(xy-y=0\) --> \(y(x-1)=0\) --> \(y=0\) or \(x=1\).

Please share your Algebra tips below and get kudos point. Thank you.
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