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Solving Quadratic Equations Quiz 2

mathematics

 What must be added to the equation ${{y}^{2}}-9y=0$ so that it has double roots?

 Find a fraction such that it forms the double roots of the given equation.

 $\begin{align} & \text{Using the concept of completing the square method}.\\ & {{y}^{2}}-9y=0\\ &\Rightarrow {{y}^{2}}-9y+{{\left( \dfrac{9}{2} \right)}^{2}}=0\\ &\Rightarrow {{\left( y-\dfrac{9}{2} \right)}^{2}}=0\\ &\Rightarrow y=\dfrac{9}{2},\dfrac{9}{2}\end{align}$

 The length of a wall is $4$m more than its breadth. A room is enclosed by four such walls. The sum of the surface area of one wall and 32 is 11 times the length of the wall. Avan wants to paint all the walls of the room such that he spends a minimum amount. If the cost of painting $1{{m}^{2}}$ is Rs. 50, then what would be the total cost for painting the room? Solve the equation using the method of factorisation.

 Find the dimensions of one wall by solving the quadratic equation formed. Then multiply it with 4 and Rs. 50 to get the total cost.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5074b985-2777-4962-843c-e374aa87dae56707768830449181444.png"/>$\begin{align} & \text{The surface area of one wall} = x\left( x+4 \right)\\ & \text{Applying the condition, we get} \\ & \Rightarrow x\left( x+4 \right)+32=11\left( x+4 \right)\\ & \text{On simplifying, we get};\\ & \Rightarrow {{x}^{2}}-7x+12=0\\ & \Rightarrow {{x}^{2}}-3x-4x+12=0\\ & \Rightarrow \left( x-3 \right) \text{and} \left( x-4 \right)=0\\ & \therefore x=3,4\\ & \text{To get the least cost take} x = 3\\ & \text{Therefore, for breadth as } 3m,\\ & \text{length} = 3 + 4 = 7m\\ & \text{Total surface area of room} = 4\times 3\times 7=84{{m}^{2}}\\ & \text{Total cost of painting the room} = 84\times 50=Rs.4200 \end{align}$ 

 The product of roots of a quadratic equation is 44 and the sum of its roots is 12. Find the quadratic equation.

 Recollect that the sum of roots is $-\dfrac{b}{a}$ and the product of roots is $\dfrac{c}{a}$.

 $\begin{align} & \text{Let us take the roots as} “a” \text{and} “b”.\\ & \text{The product of roots} = ab = 44.\\ & \text{The sum of roots} = a + b = 12.\\ & \text{So, we have} \Rightarrow a\left( 12-a \right)=44\\ & \Rightarrow {{a}^{2}}-12a+44=0\end{align}$

 What is the solution for the equation $5{{s}^{2}}-4s+3=0$ ?

 Use the quadratic formula to solve the equation.

 $\begin{align} & \text{Using quadratic formula, we get}\\ & 5{{s}^{2}}-4s+3=0\\ & \Rightarrow s=\dfrac{-(-4)\pm \sqrt{{{(-4)}^{2}}-4\times 5\times 3}}{2\times 5}\\ & \Rightarrow s=\dfrac{4\pm \sqrt{-44}}{10}\\ & \Rightarrow s=\dfrac{4\pm 2i\sqrt{11}}{10}\\ & \Rightarrow s=\dfrac{2\pm \sqrt{11}i}{5}\end{align}$

 The sum of an infinite G.P is 2 and the sum of cubes of its terms is 46. Is this series possible? If yes, then what is the common ratio of original G.P?

 Use the formula for the sum of infinite G.P $a,ar,a{{r}^{2}},.....$ as $S=\dfrac{a}{1-r}$.

 $\begin{align} & \text{We know that the sum of G.P is 2, so for a G.P } a,ar,a{{r}^{2}},..... \text{ we have}\\ &  \dfrac{a}{1-r}=2\\ & a=2\left( 1-r \right)\\ & \text{Now, sum of cube of its terms, i.e. G.P series } {{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}},..... is 46.\\ & \text{So},  \dfrac{{{a}^{3}}}{1-{{r}^{3}}}=46\\ & \because a=2\left( 1-r \right)\\ & \Rightarrow {{\left( 2\left( 1-r \right) \right)}^{3}}=46\left( 1-{{r}^{3}} \right)\\ & \Rightarrow 8{{\left( 1-r \right)}^{3}}=46\left( 1-r \right)\left( 1+r+{{r}^{2}} \right)\\ & \Rightarrow 4+4{{r}^{2}}-8r=23+23r+23{{r}^{2}}\\ & \Rightarrow 19{{r}^{2}}+31r+19=0\\ & \Rightarrow r=\dfrac{-31\pm \sqrt{{{\left( -31 \right)}^{2}}-4\times 19\times 19}}{2\times 19}\\ & \Rightarrow r=\dfrac{-31\pm \sqrt{-483}}{38}\\ & \Rightarrow r=\dfrac{-31\pm \sqrt{483}i}{38}\\ & \text{The value of} “r” \text{is imaginary and so the series is not possible}\end{align}$ 