# Tìm hiểu Intermediate Algebra – Solve Using The Square Root Property (3X+1)^2=16

## Step 1 :

1.1 Evaluate : (3x-1)2 = 9×2-6x+1

## Step 2 :

Pulling out like terms :2.1 Pull out like factors:9×2 – 6x – 15=3•(3×2 – 2x – 5)Trying to factor by splitting the middle term

2.2Factoring 3×2 – 2x – 5 The first term is, 3×2 its coefficient is 3.The middle term is, -2x its coefficient is -2.The last term, “the constant”, is -5Step-1 : Multiply the coefficient of the first term by the constant 3•-5=-15Step-2 : Find two factors of -15 whose sum equals the coefficient of the middle term, which is -2.

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 -15 + 1 = -14 -5 + 3 = -2 That”s it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -5 and 33×2 – 5x+3x – 5Step-4 : Add up the first 2 terms, pulling out like factors:x•(3x-5) Add up the last 2 terms, pulling out common factors:1•(3x-5) Step-5:Add up the four terms of step4:(x+1)•(3x-5)Which is the desired factorization

Equation at the end of step 2 :

3 • (3x – 5) • (x + 1) = 0

## Step 3 :

Theory – Roots of a product :3.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

Equations which are never true:3.2Solve:3=0This equation has no solution. A a non-zero constant never equals zero.

Solving a Single Variable Equation:3.3Solve:3x-5 = 0Add 5 to both sides of the equation:3x = 5 Divide both sides of the equation by 3:x = 5/3 = 1.667

Solving a Single Variable Equation:3.4Solve:x+1 = 0Subtract 1 from both sides of the equation:x = -1

### Supplement : Solving Quadratic Equation Directly

Solving 3×2-2x-5 = 0 directly Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex:4.1Find the Vertex ofy = 3×2-2x-5Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting “y” because the coefficient of the first term,3, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the x-coordinate of the vertex is given by -B/(2A). In our case the x coordinate is 0.3333Plugging into the parabola formula 0.3333 for x we can calculate the y-coordinate:y = 3.0 * 0.33 * 0.33 – 2.0 * 0.33 – 5.0 or y = -5.333

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = 3×2-2x-5 Axis of Symmetry (dashed) {x}={ 0.33} Vertex at {x,y} = { 0.33,-5.33} x-Intercepts (Roots) : Root 1 at {x,y} = {-1.00, 0.00} Root 2 at {x,y} = { 1.67, 0.00}

Solve Quadratic Equation by Completing The Square

4.2Solving3x2-2x-5 = 0 by Completing The Square.Divide both sides of the equation by 3 to have 1 as the coefficient of the first term :x2-(2/3)x-(5/3) = 0Add 5/3 to both side of the equation : x2-(2/3)x = 5/3Now the clever bit: Take the coefficient of x, which is 2/3, divide by two, giving 1/3, and finally square it giving 1/9Add 1/9 to both sides of the equation :On the right hand side we have:5/3+1/9The common denominator of the two fractions is 9Adding (15/9)+(1/9) gives 16/9So adding to both sides we finally get:x2-(2/3)x+(1/9) = 16/9Adding 1/9 has completed the left hand side into a perfect square :x2-(2/3)x+(1/9)=(x-(1/3))•(x-(1/3))=(x-(1/3))2 Things which are equal to the same thing are also equal to one another. Sincex2-(2/3)x+(1/9) = 16/9 andx2-(2/3)x+(1/9) = (x-(1/3))2 then, according to the law of transitivity,(x-(1/3))2 = 16/9We”ll refer to this Equation as Eq.

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#4.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x-(1/3))2 is(x-(1/3))2/2=(x-(1/3))1=x-(1/3)Now, applying the Square Root Principle to Eq.#4.2.1 we get:x-(1/3)= √ 16/9 Add 1/3 to both sides to obtain:x = 1/3 + √ 16/9 Since a square root has two values, one positive and the other negativex2 – (2/3)x – (5/3) = 0has two solutions:x = 1/3 + √ 16/9 orx = 1/3 – √ 16/9 Note that √ 16/9 can be written as√16 / √9which is 4 / 3

4.3Solving3x2-2x-5 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B and C are numbers, often called coefficients, is given by :-B± √B2-4ACx = ————————2A In our case,A= 3B= -2C= -5 Accordingly,B2-4AC=4 – (-60) = 64Applying the quadratic formula : 2 ± √ 64 x=—————6Can √ 64 be simplified ?Yes!The prime factorization of 64is2•2•2•2•2•2 To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).√ 64 =√2•2•2•2•2•2 =2•2•2•√ 1 =±8 •√ 1 =±8 So now we are looking at:x=(2±8)/6Two real solutions:x =(2+√64)/6=(1+4)/3= 1.667 or:x =(2-√64)/6=(1-4)/3= -1.000

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