# Chapterwise Topicwise Solved Papers Mathematics For Engineering Entrances 2020

### Reformatting the input đầu vào :

Changes made to your input should not affect the solution: (1): "x2" was replaced by "x^2". 1 more similar replacement(s).

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## Step 1 :

Trying lớn factor by splitting the middle term

1.1Factoring x2-x+1 The first term is, x2 its coefficient is 1.The middle term is, -x its coefficient is -1.The last term, "the constant", is +1Step-1 : Multiply the coefficient of the first term by the constant 1•1=1Step-2 : Find two factors of 1 whose sum equals the coefficient of the middle term, which is -1.

 -1 + -1 = -2 1 + 1 = 2

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Trying to factor by splitting the middle term

1.2Factoring x2-x+2 The first term is, x2 its coefficient is 1.The middle term is, -x its coefficient is -1.The last term, "the constant", is +2Step-1 : Multiply the coefficient of the first term by the constant 1•2=2Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is -1.

 -2 + -1 = -3 -1 + -2 = -3 1 + 2 = 3 2 + 1 = 3

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Equation at the kết thúc of step 1 :

(x2 - x + 1) • (x2 - x + 2) - 12 = 0

## Step 2 :

### Polynomial Roots Calculator :

2.1 Find roots (zeroes) of : F(x) = x4-2x3+4x2-3x-10Polynomial Roots Calculator is a phối of methods aimed at finding values ofxfor which F(x)=0 Rational Roots chạy thử is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integersThe Rational Root Theorem states that if a polynomial zeroes for a rational numberP/Q then p is a factor of the Trailing Constant and Q is a factor of the Leading CoefficientIn this case, the Leading Coefficient is 1 & the Trailing Constant is -10. The factor(s) are: of the Leading Coefficient : 1of the Trailing Constant : 1 ,2 ,5 ,10 Let us test ....

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PQP/QF(P/Q)Divisor
-11 -1.00 0.00x+1
-21 -2.00 44.00
-51 -5.00 980.00
-101-10.0012420.00
11 1.00 -10.00
21 2.00 0.00x-2
51 5.00 450.00
101 10.00 8360.00

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p chú ý that q and phường originate from P/Q reduced to its lowest terms In our case this means that x4-2x3+4x2-3x-10can be divided by 2 different polynomials,including by x-2

### Polynomial Long Division :

2.2 Polynomial Long Division Dividing : x4-2x3+4x2-3x-10("Dividend") By:x-2("Divisor")

 dividend x4 - 2x3 + 4x2 - 3x - 10 -divisor * x3 x4 - 2x3 remainder 4x2 - 3x - 10 -divisor * 0x2 remainder 4x2 - 3x - 10 -divisor * 4x1 4x2 - 8x remainder 5x - 10 -divisor * 5x0 5x - 10 remainder 0

Quotient : x3+4x+5 Remainder: 0

### Polynomial Roots Calculator :

2.3 Find roots (zeroes) of : F(x) = x3+4x+5See theory in step 2.1 In this case, the Leading Coefficient is 1 and the Trailing Constant is 5. The factor(s) are: of the Leading Coefficient : 1of the Trailing Constant : 1 ,5 Let us chạy thử ....

PQP/QF(P/Q)Divisor
-11 -1.00 0.00x+1
-51 -5.00 -140.00
11 1.00 10.00
51 5.00 150.00

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p cảnh báo that q and p. Originate from P/Q reduced khổng lồ its lowest terms In our case this means that x3+4x+5can be divided with x+1

### Polynomial Long Division :

2.4 Polynomial Long Division Dividing : x3+4x+5("Dividend") By:x+1("Divisor")

 dividend x3 + 4x + 5 -divisor * x2 x3 + x2 remainder - x2 + 4x + 5 -divisor * -x1 - x2 - x remainder 5x + 5 -divisor * 5x0 5x + 5 remainder 0

Quotient : x2-x+5 Remainder: 0

Trying to lớn factor by splitting the middle term

2.5Factoring x2-x+5 The first term is, x2 its coefficient is 1.The middle term is, -x its coefficient is -1.The last term, "the constant", is +5Step-1 : Multiply the coefficient of the first term by the constant 1•5=5Step-2 : Find two factors of 5 whose sum equals the coefficient of the middle term, which is -1.

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 -5 + -1 = -6 -1 + -5 = -6 1 + 5 = 6 5 + 1 = 6

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Equation at the over of step 2 :

(x2 - x + 5) • (x + 1) • (x - 2) = 0

## Step 3 :

Theory - Roots of a hàng hóa :3.1 A hàng hóa of several terms equals zero.When a hàng hóa 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 lớn solve as many equations as there are terms in the productAny solution of term = 0 solves hàng hóa = 0 as well.

Parabola, Finding the Vertex:3.2Find the Vertex ofy = x2-x+5Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up và 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 lớn 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 0.5000Plugging into the parabola formula 0.5000 for x we can calculate the y-coordinate:y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 + 5.0 or y = 4.750

Parabola, Graphing Vertex & X-Intercepts :

Root plot for : y = x2-x+5 Axis of Symmetry (dashed) x= 0.50 Vertex at x,y = 0.50, 4.75 Function has no real roots

Solve Quadratic Equation by Completing The Square

3.3Solvingx2-x+5 = 0 by Completing The Square.Subtract 5 from both side of the equation :x2-x = -5Now the clever bit: Take the coefficient of x, which is 1, divide by two, giving 1/2, và finally square it giving 1/4Add 1/4 to lớn both sides of the equation :On the right hand side we have:-5+1/4or, (-5/1)+(1/4)The common denominator of the two fractions is 4Adding (-20/4)+(1/4) gives -19/4So adding to both sides we finally get:x2-x+(1/4) = -19/4Adding 1/4 has completed the left hand side into a perfect square :x2-x+(1/4)=(x-(1/2))•(x-(1/2))=(x-(1/2))2 Things which are equal khổng lồ the same thing are also equal to lớn one another. Sincex2-x+(1/4) = -19/4 andx2-x+(1/4) = (x-(1/2))2 then, according to lớn the law of transitivity,(x-(1/2))2 = -19/4We"ll refer to this Equation as Eq. #3.3.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x-(1/2))2 is(x-(1/2))2/2=(x-(1/2))1=x-(1/2)Now, applying the Square Root Principle khổng lồ Eq.#3.3.1 we get:x-(1/2)= √ -19/4 Add 1/2 to both sides khổng lồ obtain:x = một nửa + √ -19/4 In Math,iis called the imaginary unit. It satisfies i2=-1. Both i & -i are the square roots of -1Since a square root has two values, one positive and the other negativex2 - x + 5 = 0has two solutions:x = 50% + √ 19/4 • iorx = một nửa - √ 19/4 • iNote that √ 19/4 can be written as√19 / √4which is √19 / 2

### Solve Quadratic Equation using the Quadratic Formula

3.4Solvingx2-x+5 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B và C are numbers, often called coefficients, is given by :-B± √B2-4ACx = ————————2A In our case,A= 1B= -1C= 5 Accordingly,B2-4AC=1 - đôi mươi =-19Applying the quadratic formula : 1 ± √ -19 x=—————2In the mix of real numbers, negative numbers vì chưng not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)Both i và -i are the square roots of minus 1Accordingly,√-19=√19•(-1)=√19•√-1=±√ 19 •i √ 19 , rounded khổng lồ 4 decimal digits, is 4.3589So now we are looking at:x=(1± 4.359 i )/2Two imaginary solutions :

x =(1+√-19)/2=(1+i√ 19 )/2= 0.5000+2.1794ior: x =(1-√-19)/2=(1-i√ 19 )/2= 0.5000-2.1794iSolving a Single Variable Equation:3.5Solve:x+1 = 0Subtract 1 from both sides of the equation:x = -1

Solving a Single Variable Equation:3.6Solve:x-2 = 0Add 2 to both sides of the equation:x = 2

## Four solutions were found :

x = 2x = -1x =(1-√-19)/2=(1-i√ 19 )/2= 0.5000-2.1794ix =(1+√-19)/2=(1+i√ 19 )/2= 0.5000+2.1794i