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This is the second of our many quadratic equation quizzes. Click on this to access more such competitive quizzes on quadratics.

Quadratics Quiz 2

mathematics

 Choose the option which does not match with the definition of a quadratic equation.

 A quadratic equation must have a degree of 2 or the highest power of the variable in the equation must be 2. Use this fact to check each option and conclude the right option.

 $\begin{align} & \text{First convert all the options in the form } a{{x}^{2}}+bx+c=0.\\ & \text{On Comparing all the simplified forms}.\\ & \text{Option C, results in a linear equation, because the highest power of the variable is 1}. \\ & {{\left( 1-y \right)}^{2}}-{{\left( 1+y \right)}^{2}}=2 \\ & \Rightarrow 1+{{y}^{2}}-2y-1-{{y}^{2}}-2y=2 \\ & \Rightarrow -4y=2 \\ & \Rightarrow -2y=1.\\ & \text{We know that for a quadratic equation, the highest power of the variable must be 2}.\\ & \text{So, the only option which doesn’t match with the definition of a quadratic equation is option C.} \end{align}$

 Identify the incorrect statement from the below given options.

 The roots of a quadratic equation can be real as well as complex, having both positive and negative roots. They also have roots that have the same magnitude which means double root.

 On recollecting the facts about the nature of roots of the quadratic equation we can say that the statement of option $B$ is not true. Because, the word atmost indicates that a quadratic equation can have 0, 1 or 2 roots. But, it is known that there must be exactly two roots. 

 What is the term ${{b}^{2}}-4ac$ for the quadratic equation $a{{x}^{2}}+bx+c=0$ called?

 The term ${{b}^{2}}-4ac$ is often represented by the letter D and can be useful for determining the nature of roots.

 We know that the term ${{b}^{2}}-4ac$ is called the discriminant of the quadratic equation $a{{x}^{2}}+bx+c=0$. It helps in determining the nature of the roots. The usual notation is $D={{b}^{2}}-4ac$. Depending on the value of the discriminant, the nature of roots can be real, imaginary, equal or unequal roots. For example, if $D > 0$, then the roots are said to be real and unequal.

 Come up with a quadratic equation to represent the below situation mathematically.<br>We have to find the area of a quadrilateral ABCD as shown in the figure below. It is made up of a rectangle AECF and a square EFBD. The length of the rectangle is $5 cm$. If the area of the quadrilateral is $20cm^{2}$ , then obtain a quadratic equation to find the breadth of the rectangle. <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/21e2855e-6e75-42c1-b068-1933cd36730e8594396977189681249.png"/>

 First, assume breadth of rectangle as b cm. Use the fact that the area of the rectangle is the product of length and breadth and the area of the square is square of its side. Add them up and equate to the given area to get the required quadratic equation.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7074112c-e83b-49c0-9cf0-b5bd558046d1685319677016243845.png"/>$\begin{align} & \text{From the quadrilateral ABCD. Area of ABCD = Area of ACEF + Area of EFBD} --(1),\\ & \text{where Area of ABCD is } 20cm^{2}.\\ & \text{To find the breadth of the rectangle}, \\ & \text{Consider } AC = bcm.\\ & \text{Since the breadth of the rectangle is equal to the side of the square, then } AC = EF = b. \\ & \text{Area of rectangle} = \text{length} \times \text{breadth} \\ & \Rightarrow CF\times AC \\ & \Rightarrow 5\times b \\ & \text{Area of square }= {side}^{2} \\ & \Rightarrow E{{F}^{2}} \\ & \Rightarrow {{b}^{2}}.\\ & \text{Substituting the above in eq(1), we get}.\\ & \text{Area of ABCD }=5b+{{b}^{2}}\\ & \text{W.K.T area of ABCD}= 20cm^{2}.\\ & \text{On equating the value and rearranging it, we get}\\ & \Rightarrow {{b}^{2}}+5b-20=0\\ & \text{Hence, the QE of the breadth of the rectangle is} {{b}^{2}}+5b-20=0\end{align}$ 

 $x=\dfrac{1}{2}$ can be a root of which of the below equations?

 The value of $x=\dfrac{1}{2}$  must be substituted in the given equations. The equation in which the LHS = RHS = 0 would be the required quadratic equation.

 $\begin{align} & \text{On substituting the value of }x=\dfrac{1}{2}\text{ in }8{{x}^{2}}+2x-3=0,\text{ we get }\\ & \Rightarrow 8{{\left( \dfrac{1}{2} \right)}^{2}}+2\left( \dfrac{1}{2} \right)-3=0 \\ & \Rightarrow 8\times \dfrac{1}{4}+1-3=0 \\ & \Rightarrow 2+1-3=0 \\ & \Rightarrow 0=0 \end{align}$ 