What are the solutions to the equation x2 6x 40?

Answer

Table of Contents Show

Step by step solution :
Step 1 :
Step 2 :
Supplement : Solving Quadratic Equation Directly
Solve Quadratic Equation using the Quadratic Formula
Two solutions were found :

Verified

Hint: Here given an algebraic equation firstly we have to convert the given equation to the quadratic equation \[a{x^2} + bx + c = 0\] by shifting 40 from RHS to LHS. Later solve the quadratic equation by using the method of factorisation and find the factors and equate each factor to the zero to get the value of \[x\] .

Complete step-by-step answer:

Factorization means the process of creating list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\] . Here, b is the sum of d and e and c is the product of d and e. Consider the given expression: \[ \Rightarrow {x^2} + 6x = 40\] Subtract both side by 40, then \[ \Rightarrow {x^2} + 6x - 40 = 40 - 40\] On simplification, we get \[ \Rightarrow {x^2} + 6x - 40 = 0\] The above equation is similar to a quadratic equation \[a{x^2} + bx + c\] now, solved by the method of factorization.Now, Break the middle term as the summation of two numbers such that its product is equal to -40. Calculated above such two numbers are -4 and 10. \[ \Rightarrow {x^2} - 4x + 10x - 40 = 0\] Making pairs of terms in the above expression \[ \Rightarrow \left( {{x^2} - 4x} \right) + \left( {10x - 40} \right) = 0\] Take out greatest common divisor GCD from the both pairs, then \[ \Rightarrow x\left( {x - 4} \right) + 10\left( {x - 4} \right) = 0\] Take \[\left( {x - 4} \right)\] common \[ \Rightarrow \left( {x - 4} \right)\left( {x + 10} \right) = 0\] Equate the each factor to zero, then \[ \Rightarrow \left( {x - 4} \right) = 0\] or \[\left( {x + 10} \right) = 0\] \[ \Rightarrow x - 4 = 0\] OR \[x + 10 = 0\] \[ \Rightarrow x = 4\] OR \[x = - 10\] Hence, the required solution is \[x = 4\] or \[x = - 10\] .

So, the correct answer is “\[x = 4\] or \[x = - 10\]”.

Note: The equation is a quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\] , the product of \[a{x^2}\] and \[c\] is equal to the sum of \[bx\] of the equation. Hence, we obtain the factors. The factors for the equation depends on degree of the equation and this type of quadratic equation also solve by using completing the square method and also by using a quadratic formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1 Factoring x2-6x-40

The first term is, x2 its coefficient is 1 .

The middle term is, -6x its coefficient is -6 .
The last term, "the constant", is -40

Step-1 : Multiply the coefficient of the first term by the constant 1 • -40 = -40

Step-2 : Find two factors of -40 whose sum equals the coefficient of the middle term, which is -6 .

-40	+	1	=	-39
-20	+	2	=	-18
-10	+	4	=	-6	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and 4
x2 - 10x + 4x - 40Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-10)

Add up the last 2 terms, pulling out common factors :

4 • (x-10)

Step-5 : Add up the four terms of step 4 :

(x+4) • (x-10)

Which is the desired factorization

Equation at the end of step 1 :

(x + 4) • (x - 10) = 0

Step 2 :

Theory - Roots of a product :

2.1 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.

Solving a Single Variable Equation :

2.2 Solve : x+4 = 0Subtract 4 from both sides of the equation :
x = -4

Solving a Single Variable Equation :

2.3 Solve : x-10 = 0Add 10 to both sides of the equation :
x = 10

Supplement : Solving Quadratic Equation Directly

Solving x2-6x-40 = 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 :

3.1      Find the Vertex of   Parabolas 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, 1 , 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 3.0000 Plugging into the parabola formula 3.0000 for x we can calculate the y -coordinate :
  y = 1.0 * 3.00 * 3.00 - 6.0 * 3.00 - 40.0
or   y = -49.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-6x-40
Axis of Symmetry (dashed) {x}={ 3.00}
Vertex at {x,y} = { 3.00,-49.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-4.00, 0.00}
Root 2 at {x,y} = {10.00, 0.00}

Solve Quadratic Equation by Completing The Square

3.2 Solving x2-6x-40 = 0 by Completing The SquareAdd 40 to both side of the equation :
x2-6x = 40

Now the clever bit: Take the coefficient of x , which is 6 , divide by two, giving 3 , and finally square it giving 9

Add 9 to both sides of the equation :

On the right hand side we have :

40 + 9 or, (40/1)+(9/1)

The common denominator of the two fractions is 1 Adding (40/1)+(9/1) gives 49/1 So adding to both sides we finally get :

x2-6x+9 = 49

Adding 9 has completed the left hand side into a perfect square :

   x2-6x+9  =
   (x-3) • (x-3)  =
  (x-3)2 Things which are equal to the same thing are also equal to one another. Since

x2-6x+9 = 49 and

x2-6x+9 = (x-3)2 then, according to the law of transitivity,

(x-3)2 = 49

We'll refer to this Equation as Eq. #3.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-3)2 is

   (x-3)2/2 =
  (x-3)1 =
   x-3

Now, applying the Square Root Principle to Eq. #3.2.1 we get:

x-3 = √ 49

Add 3 to both sides to obtain:

x = 3 + √ 49 Since a square root has two values, one positive and the other negative

x2 - 6x - 40 = 0

has two solutions:

x = 3 + √ 49

x = 3 - √ 49

Solve Quadratic Equation using the Quadratic Formula

3.3 Solving x2-6x-40 = 0 by the Quadratic FormulaAccording to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC x = ———————— 2A

In our case, Accordingly,

B2 - 4AC = 36 - (-160) = 196

Applying the quadratic formula :

Can √ 196 be simplified ?

Yes! The prime factorization of 196 is

2•2•7•7
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).

√ 196   =  √ 2•2•7•7   =2•7•√ 1   =
                ±  14 • √ 1   =
                ±  14

So now we are looking at:

x = ( 6 ± 14) / 2

Two real solutions:

x =(6+√196)/2=3+7= 10.000