# Algebra II Step-by-Step Solving Problems

The main topic in Algebra 2 is quadratic equation and its graph named parabola. When you want to solve quadratic equation you need to know how to factor trinomials. In Algebra 2 you will learn how to solve quadratic equations by factoring, completing the square and using the quadratic formula

## Things you’ll need

Algebra 2 textbook

## Solving Algebra 2 problems step-by-step

How to Solve Quadratic Equations Like 3x^2 + 5x + 3 = x(3 — x) — 6x By Factoring

– You must simplify the equation by clearing parentheses and combining like terms. Use distributive property to clear parentheses. In this case, clear parentheses by multiplying every term in (3-x) by X.

3x^2 + 5x + 3 = 3x — x^2 — 6x

Combine like terms: 3x^2 + 5x + 3 = –3x — x^2

– Arrange the terms into standard quadratic form. You need to add and subtract terms from the one side to another side. Variable term with the exponent of 2 is written first and variable with the exponent of 1 is written next. Variables with no exponents is written next to it and is understood to have exponent of 1. Constant term is written last. Other side of the equation will be equal to 0.

Add x^2 to both sides:3x^2 + x^2 + 5x + 3 = — 3x — x^2 + x^2

Now we have 4x^2 + 5x + 3 = — 3x.

Add 3x to both sides: 4x^2 + 5x + 3x + 3 = — 3x + 3x

Now we have 4x^2 + 8x + 3 = 0.

Note that the square term, 4x^2, comes first. The first-power term, 8x, is second. The constant is written after all.

– Now you need to factor equation completely.

(2x + 1)(2x + 3) = 0.

– Set each factor equal to 0 and solve for x.

2x + 1 = 0 and 2x + 3 = 0

When we solve first equation, we get x = — 1/2.
When we solve second equation, we get  x = — 3/2.

This is solution to quadratic equation.

How to Solve Quadratic Equations Like 4x^2 + 8x + 3 = 0 by Completing the Square

– You need to move the constant term to the next side, and you will do that using addition or subtraction. In this example 3 has been added to the left side, and we do the opposite, which is subtracting 3. It is important to subtract 3 from the both sides of the equation because what is done on the one side, must be done on the another side.

The equation becomes 4x^2 + 8x = — 3

– Now you need to factor out constant of the square term.

4(x^2 + 2x) = — 3

– Take half the coefficient of the first-power term (the 2x) and square it. The coefficient is the number in front of the variable. The number in front of the first-power term is 2. Half of 2 is 1 and 1^2 is equal to 1.

– The value that you found must be inserted squaring half the coefficient of the first-power term into the parentheses:

4(x^2 + 2x + 1)

Note that x^2 + 2x + 1 is a perfect square trinomial. When you complete the square, the result will be a perfect square trinomial, which can always be factored into the square of a binomial.

– Whatever you have added to the left side, you must add to right side, too.

You have not added 1 to the left side of the equation because that 1 is in parentheses and the terms inside parentheses have been multiplied by 4, you have actually added 4 to the left side. Therefore, you must add 4 to the right side.

The equation now becomes 4(x^2 + 2x + 1) = — 3 + 4, which simplifies to 4(x^2 + 2x + 1) = 1

See how  x^2 + 2x + 1 is a perfect square trinomial. The process of completing the square will always give a perfect square trinomial which can always be factored into the square of a binomial.

– Factor the expression inside parentheses

Now we have 4(x + 1)^2 = 1.

This expression (x + 1)^2 is the square of a binomial.

– Now we must divide both sides by 4 to get the square term by itself.

(x + 1)^2 = 1/4

– Take the square root of both sides.

Thus, x + 1 = +/– sqrt(1/4)

The square root of 1/4 is either + 1/2 or — 1/2

– Set the expression in parentheses to be equal to the values of the term on the right side  and solve it. Remember that you will need two equations, one for each value.

x + 1 = 1/2

x + 1 = — 1/2

Solve each equation separately and find the answers

x = — 1/2 and x = — 3/2.

These are the same answers we obtained by factoring.

Solve Equations Like x^2 — 8x = 9 using the quadratic formula.

– Put the terms of the equation in regular quadratic form — in descending order of the variables with the equation set equal to 0. In this equation, subtract 9 from both sides to make the equation equal to 0. Subtracting 9 gives x^2 — 8x — 9 = 0.

– Identify the coefficients a, b, and c. The letter A represents the coefficient of the square term, b represents the coefficient of the first power term and C represents the constant.

We got:  a = 1, b = — 8, and c = — 9.

– Substitute the values for a, b, and c into the quadratic formula.

The quadratic formula is x = [– b +/– sqrt(b^2 — 4ac)]/2a.

In this problem, substituting produces x = [– (–8) +/–sqrt((– 8)^2 — 4(1)( — 9))]/2(1)

-Perform appropriate arithmetic operation to simplify the expression.

When you do simplifying always simplify what is in grouping symbols, like parentheses, brackets and radical signs. Do it first. If some expression contains more then one set of grouping symbols you need to work with the innermost grouping symbol fist. For this instance, you should work with what is under the radical.

See that — (– 8) is equal to 8.

x = [8 +/– sqrt(64 + 36)]/2

x = [8 +/– sqrt(100)]/2

x = [8 +/– 10]/2

When you perform the computations it gives two values for x, 9 and –1

These are the solutions to the equation. These values also indicate where the graph of the quadratic equation, a parabola, crosses the x-axis: at the point (9,0) and (–1,0). These are often called the zeros of the equation.

X = (8 — 10)/2 = –1