Reformatting the input :

Changes made to her input must not influence the solution: (1): "x2" was changed by "x^2".

You are watching: 2x^2+2x^2

Step by action solution :

Step 1 :

Equation at the end of action 1 :

(2x2 - 2x) - 2 = 0

Step 2 :

Step 3 :

Pulling out favor terms :3.1 traction out favor factors:2x2 - 2x - 2=2•(x2 - x - 1)

Trying to element by separating the middle term

3.2Factoring x2 - x - 1 The an initial term is, x2 the coefficient is 1.The center term is, -x that is coefficient is -1.The last term, "the constant", is -1Step-1 : main point the coefficient that the very first term by the continuous 1•-1=-1Step-2 : find two determinants of -1 who sum equals the coefficient the the middle term, i m sorry is -1.

-1+1=0

Observation : No two such components can be discovered !! Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

2 • (x2 - x - 1) = 0

Step 4 :

Equations i beg your pardon are never true:4.1Solve:2=0This equation has actually no solution. A a non-zero constant never equates to zero.

Parabola, recognize the Vertex:4.2Find the peak ofy = x2-x-1Parabolas have a highest or a lowest allude 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 that the an initial term,1, is hopeful (greater 보다 zero).Each parabola has actually a vertical line of symmetry that passes with its vertex. Therefore symmetry, the line of the opposite would, because that example, pass v the midpoint the the 2 x-intercepts (roots or solutions) that the parabola. The is, if the parabola has actually indeed two genuine solutions.Parabolas can model many real life situations, such as the height above ground, of things thrown upward, after some period of time. The vertex of the parabola can carry out us through information, such together the maximum elevation that object, thrown upwards, can reach. Therefore we want to have the ability to find the works with of the vertex.For any type of parabola,Ax2+Bx+C,the x-coordinate of the vertex is offered by -B/(2A). In our case the x coordinate is 0.5000Plugging into the parabola formula 0.5000 for x we have the right to calculate the y-coordinate:y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 - 1.0 or y = -1.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot because that : y = x2-x-1 Axis of the contrary (dashed) x= 0.50 Vertex in ~ x,y = 0.50,-1.25 x-Intercepts (Roots) : root 1 in ~ x,y = -0.62, 0.00 root 2 at x,y = 1.62, 0.00

Solve Quadratic Equation by completing The Square

4.3Solvingx2-x-1 = 0 by completing The Square.Add 1 to both next of the equation : x2-x = 1Now the clever bit: take the coefficient of x, which is 1, division by two, offering 1/2, and finally square it providing 1/4Add 1/4 come both political parties of the equation :On the appropriate hand side us have:1+1/4or, (1/1)+(1/4)The usual denominator that the two fractions is 4Adding (4/4)+(1/4) provides 5/4So including to both sides we lastly get:x2-x+(1/4) = 5/4Adding 1/4 has actually completed the left hand side right into a perfect square :x2-x+(1/4)=(x-(1/2))•(x-(1/2))=(x-(1/2))2 points which are equal to the same thing are additionally equal come one another.

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Sincex2-x+(1/4) = 5/4 andx2-x+(1/4) = (x-(1/2))2 then, follow to the law of transitivity,(x-(1/2))2 = 5/4We"ll refer to this Equation as Eq. #4.3.1 The Square root Principle claims that when two things space equal, your square roots room equal.Note that the square root of(x-(1/2))2 is(x-(1/2))2/2=(x-(1/2))1=x-(1/2)Now, using the Square source Principle to Eq.#4.3.1 we get:x-(1/2)= √ 5/4 add 1/2 to both political parties to obtain:x = 1/2 + √ 5/4 because a square root has actually two values, one positive and also the various other negativex2 - x - 1 = 0has two solutions:x = 1/2 + √ 5/4 orx = 1/2 - √ 5/4 note that √ 5/4 can be created as√5 / √4which is √5 / 2

Solve Quadratic Equation using the Quadratic Formula

4.4Solvingx2-x-1 = 0 by the Quadratic Formula.According come the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , whereby A, B and C space numbers, often called coefficients, is given by :-B± √B2-4ACx = ————————2A In ours case,A= 1B= -1C= -1 Accordingly,B2-4AC=1 - (-4) = 5Applying the quadratic formula : 1 ± √ 5 x=————2 √ 5 , rounded to 4 decimal digits, is 2.2361So currently we are looking at:x=(1± 2.236 )/2Two actual solutions:x =(1+√5)/2= 1.618 or:x =(1-√5)/2=-0.618