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Subtracting polynomials Quiz 2

mathematics

 The number of items present in a self, which contains only books and notebooks is Rs. 100. The books are 80 more than the notebooks, find the number of notebooks in that self.

 Form two equations with the relations given then solve them.

  $\text{Let the number of books and notebooks be $x$ and $y$ respectively.} \\ \text{Now, } x+y=100\ldots (1) \\  x-y=80\ldots (2) \\ \text{Solving both the equations, }y=10.$

 The roots of a quadratic equation $P{{\left( x - 2 \right)}^{2}}Q{{\left( x - 5 \right)}^{2}}2=0$ are real and identical. If $P:Q=3:2$ , find $P$.

 Expand the equations then form a quadratic equation for applying the discriminant rule, from there using the ratio given find $P$.

  $P{{\left( x-2 \right)}^{2}}~-Q{{\left( x-5 \right)}^{2}}~-2=0 \\  P[{{x}^{2}}~-4x+4\left] -Q \right[{{x}^{2}}-10x+25]-2=0 \\  \left( P-Q \right){{x}^{2}}+\left( 10Q-4P \right)x+\left( 4P-25Q-2 \right)=0 \\  \text{As the roots are real and identical, the value of the discriminant equal to zero}\text{.i}\text{.e}\text{. }{{b}^{2}}~-4ac=0 \\  {{\left( 10Q-4P \right)}^{2}}-4\left( P-Q \right)\left( 4P-25Q-2 \right)=0 \\  \text{Given that }P:Q=3:2 \\  \text{Let }P=3k\text{ and }Q=2k. \\  \text{Then, }{{\left[ 10\left( 2k \right)-4\left( 3k \right) \right]}^{2}}-4\left( 3k-2k \right)\left[ 4\left( 3k \right)-25\left( 2k \right)-2 \right]=0 \\  64{{k}^{2}}~-4k\left( -38k-2 \right)=0\Rightarrow 64{{k}^{2}}+152{{k}^{2}}+8k=0 \\  216{{k}^{2}}+8k=0\Rightarrow 8k\left( 27k+1 \right)=0 \\  \text{As }k\ne 0,k=\frac{-1}{27} \\  \text{Therefore }P=3k,P=3\left( \frac{-1}{27} \right)=\frac{-1}{9} $

 If $a, b$ and $c$ are real and $\left( x - a \right)\left( x - b \right)+\left( x - b \right)\left( x - c \right)+\left( x - c \right)\left( x - a \right)=0$ if $a\ne b\ne c$ then which of the following options is true?

 Expand the equations then form a quadratic equation for applying the discriminant rule and find the required relation.

  $(x-a)\left( x-b \right)+\left( x-b \right)\left( x-c \right)+\left( x-c \right)\left( x-a \right)=0  \\   \Rightarrow {{x}^{2}}-\left( a+b \right)x+ab+{{x}^{2}}-\left( b+c \right)x+bc+{{x}^{2}}-\left( c+a \right)x+ca=0  \\   \Rightarrow 3{{x}^{2}}-2\left( a+b+c \right)x+ab+bc+ac=0  \\   \text{Discriminant of the above equation =}4{{\left( a+b+c \right)}^{2}}-12\left( ab+bc+ac \right)  \\   \Rightarrow 4\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac-3ab-3bc-3ac \right]  \\   \Rightarrow 4\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right]  \\   \Rightarrow \dfrac{4}{2}\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( a-c \right)}^{2}} \right]  \\   \Rightarrow 2\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( a-c \right)}^{2}} \right]  \\ \text{As }a,b\text{ and }c\text{ are not equal to each other as given in question, so the Discriminant of the given equation is always greater than $0$, therefore the roots are real.}$

 What is the maximum value of $\left| x+y-z \right|-z$, where $x, y$ and $z$ are real numbers, such that, $\left( x+y \right)\left( x+y+z \right)=120,\left( y+z \right)\left( x+y+z \right)=96\text{ and }\left( z+x \right)\left( x+y+z \right)=72$.

 First add the three equations form two cases once taking positive and negative RHS.

  $\text{Adding the three equations we get }2\times {{(x+y+z)}^{2}}=288 \\ \left( x+y+z \right)=\pm 12 \\ \text{Case1:} \\ \begin{array}{*{35}{l}}   x+y+z=12  \\   x+y=10,y+z=8,x+z=6\text{ which gives }\left( x,y,z \right)=\left( 4,6,2 \right)  \\   \text{Now, }\left| x+y-z \right|-z=\left| 4+6-2 \right|-2=6  \\ \end{array} \\ \text{Case2:} \\  x+y+z=-12  \\   x+y=-10,y+z=-8,x+z=-6\text{ which gives }\left( x,y,z \right)=\left( -4,-6,-2 \right)  \\   \text{Now, }\left| x+y-z \right|-z=\left| \left( -4 \right)+\left( -6 \right)-\left( -2 \right) \right|-\left( -2 \right)=8+2=10$ 

 The polynomial $f\left( x \right)=A{{x}^{39}}-B{{x}^{28}}+C{{x}^{17}}+DX+3$ leaves remainders of $4$ and $5$ on division by $\left( x-1 \right)$ and $\left( x-2 \right)$ respectively. The polynomial $g\left( x \right)=A{{x}^{39}}-B{{x}^{28}}+C{{x}^{17}}+2X+E$ is divisible by $\left( {{x}^{2}}-3x+2 \right)$. What is the value of $2D+E$?

 Use factor and remainder theorem to form relations between the unknown variables then put the values of $x$ to calculate the values $D\text{ and E}$.

  $\text{Since }g\left( x \right)\text{ is divisible by }\left( {{x}^{2}}-3x+2 \right)=\left( x-1 \right)(x-2),\text{ the factor theorem tells us that }g\left( 1 \right)=g\left( 2 \right)=0 \\ \text{We have, }f\left( x \right)=g\left( x \right)+\left( D-2 \right)x+\left( 3-E \right) \\ \text{Since, }g\left( x \right)\text{ is divisible by }\left( x-1 \right)\text{ and }\left( x-2 \right),\text{ any remainder we are going to get will be from: }\left( D-2 \right)x+\left( 3-E \right),\text{ part of }f\left( x \right) \\ \text{Let's call this }h\left( x \right)=\left( D-2 \right)x+\left( 3-E \right) \\ \text{Now, we use the remainder theorem on }h\left( x \right)=\left( D-2 \right)x+\left( 3-E \right) \\ \text{Putting }x=1,\text{ we get }h\left( 1 \right)=4=D-E+1 \\ \text{We have, }f\left( x \right)=g\left( x \right)+\left( D-2 \right)x+\left( 3-E \right) \\ \text{Since, }g\left( x \right)\text{ is divisible by }\left( x-1 \right)\text{ and }\left( x-2 \right),\text{ any remainder we are } \text{going to get will be from: }\left( D-2 \right)x+\left( 3-E \right),\text{ part of }f\left( x \right)\text{ let's call this }h(x) =  (D - 2)x + (3 - E) \\ \text{Putting }x=1,\text{ we get }h\left( 1 \right)=4=D-E+1 \\ \text{Putting }x=2,\text{ we get }h\left( 2 \right)=5=2D-E-1 \\ \text{Solving these equations} \text{ simultaneously, we get }D=3\text{ and }E=0 \\ \text{This gives }\left( 2D+E \right)=6$ 