 ## How to Solve Quadratic Equations?

What is an Equation? In mathematics, an equation is defined as a statement that has an equal “=” sign in between of two algebraic statements. Also, these two algebraic statements have equal values. The Expression on the left is called the “Left Hand Side” and the expression on the right is called the “Right Hand Side” of the equation. For an equation, the L.H.S $$=$$ R.H.S.
The algebraic expressions contained in an equation consists of both constants, variables, and mixed (variables with coefficients).

Quadratic equations are termed as those equations where the highest power or degree of the equation is 2. The standard form for a quadratic equation is $$ax^2 \ + \ bx \ + \ c \ = \ 0$$, where $$a \ ≠ \ 0$$. Now, let’s learn to solve quadratic equations by taking the example of equation $$(x \ – \ 5)(x \ + \ 3)$$.

• First, let’s apply the distributive law to simplify the equations as $$(x \ – \ 5)(x) \ + \ (x \ – \ 5)(3)$$
• By further simplification and application of distributive law, we get: $$x^2 \ - \ 5x \ + \ 3x \ – \ 15$$
• Now, let’s combine like terms to get: $$x^2 \ - \ 2x \ – \ 15$$

Example 1: Let’s solve quadratic equations by taking the example of equation $$(x \ + \ 7)(x \ - \ 5)$$.
•    First, let’s apply the distributive law to simplify the equations as: $$(x \ + \ 7)(x) \ + \ (x \ + \ 7)(-5)$$
•    By further simplification and application of distributive law we get: $$x^2 \ + \ 7x \ - \ 5x \ – \ 35$$
•    Now, let’s combine like terms to get: $$x^2 \ + \ 2x \ – \ 35$$.

Example 2: Let’s solve quadratic equations by taking the example of equation $$(x \ - \ 2)(x \ + \ 3)$$.
•    First, let’s apply the distributive law to simplify the equations as: $$(x \ - \ 2)(x) \ + \ (x \ - \ 2)(3)$$
•    By further simplification and application of distributive law we get: $$x^2 \ - \ 2x \ + \ 3x \ – \ 6$$
•    Now, let’s combine like terms to get: $$x^2 \ + \ x \ – \ 6$$

1) $$(x \ + \ 5)(x \ - \ 1) =$$

2) $$(x \ + \ 4)(x \ - \ 5) =$$

3) $$(x \ + \ 1)(x \ - \ 1) =$$

4) $$(x \ + \ 2)(x \ - \ 5) =$$

5) $$(x \ + \ 3)(x \ - \ 1) =$$

6) $$(x \ - \ 5)(x \ - \ 1) =$$

7) $$(x \ - \ 2)(x \ - \ 5) =$$

8) $$(x \ + \ 8)(x \ - \ 5) =$$

9) $$(x \ - \ 1)(x \ - \ 1) =$$

10) $$(x \ - \ 4)(x \ + \ 3) =$$

1) $$(x \ + \ 5)(x \ - \ 1) =$$$$\ \color{red}{x^2 \ + \ 5x \ - \ 1x \ + \ (5)(-1)= }$$$$\ \color{red}{x^2 \ + \ 4x \ - \ 5}$$
2) $$(x \ + \ 4)(x \ - \ 5) =$$$$\ \color{red}{x^2 \ + \ 4x \ - \ 5x \ + \ (4)(-5)= }$$$$\ \color{red}{x^2 \ - \ 1x \ - \ 20}$$
3) $$(x \ + \ 1)(x \ - \ 1) =$$$$\ \color{red}{x^2 \ + \ 1x \ - \ 1x \ + \ (1)(-1)= }$$$$\ \color{red}{x^2 \ + \ 0x \ - \ 1}$$
4) $$(x \ + \ 2)(x \ - \ 5) =$$$$\ \color{red}{x^2 \ + \ 2x \ - \ 5x \ + \ (2)(-5)= }$$$$\ \color{red}{x^2 \ - \ 3x \ - \ 10}$$
5) $$(x \ + \ 3)(x \ - \ 1) =$$$$\ \color{red}{x^2 \ + \ 3x \ - \ 1x \ + \ (3)(-1)= }$$$$\ \color{red}{x^2 \ + \ 2x \ - \ 3}$$
6) $$(x \ - \ 5)(x \ - \ 1) =$$$$\ \color{red}{x^2 \ - \ 5x \ - \ 1x \ + \ (-5)(-1)= }$$$$\ \color{red}{x^2 \ - \ 6x \ + \ 5}$$
7) $$(x \ - \ 2)(x \ - \ 5) =$$$$\ \color{red}{x^2 \ - \ 2x \ - \ 5x \ + \ (-2)(-5)= }$$$$\ \color{red}{x^2 \ - \ 7x \ + \ 10}$$
8) $$(x \ + \ 8)(x \ - \ 5) =$$$$\ \color{red}{x^2 \ + \ 8x \ - \ 5x \ + \ (8)(-5)= }$$$$\ \color{red}{x^2 \ + \ 3x \ - \ 40}$$
9) $$(x \ - \ 1)(x \ - \ 1) =$$$$\ \color{red}{x^2 \ - \ 1x \ - \ 1x \ + \ (-1)(-1)= }$$$$\ \color{red}{x^2 \ - \ 2x \ + \ 1}$$
10) $$(x \ - \ 4)(x \ + \ 3) =$$$$\ \color{red}{x^2 \ - \ 4x \ + \ 3x \ + \ (-4)(3)= }$$$$\ \color{red}{x^2 \ - \ 1x \ - \ 12}$$

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