Answer
Verified
407.1k+ views
Hint:Assume that the quadratic equations are $ \left( x-a \right)\left( x-1 \right) $ and $ \left( x-b \right)\left( x-1 \right) $ . Expand each of these expressions and find the discriminant of both the quadratic expressions. Use the fact that the discriminant of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given by $ D={{b}^{2}}-4ac $ .Equate the two discriminants and hence find the relation between a and b. Hence find which of the options are correct and which options are incorrect.
Complete step-by-step answer:
Let the other root of the first expression be a and the other root of the second expression be b
Since both the equations have x =1 as one of their roots, we have the quadratic expressions are
$ \left( x-1 \right)\left( x-a \right)=0\text{ and }\left( x-1 \right)\left( x-b \right)=0 $
Expanding $ \left( x-1 \right)\left( x-a \right) $
We know that $ \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab $
Hence, we have
$ \left( x-1 \right)\left( x-a \right)={{x}^{2}}-\left( a+1 \right)x+a $
We know that that the discriminant of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given by $ D={{b}^{2}}-4ac $
Here a = 1, b = -(a+1) and c = a
Hence the discriminant of the expression is $ {{D}_{1}}={{\left( a+1 \right)}^{2}}-4a $
Using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ , we get
$ {{D}_{1}}={{a}^{2}}+1+2a-4a={{a}^{2}}-2a+1 $
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $
Hence, we have
$ {{D}_{1}}={{\left( a-1 \right)}^{2}} $
Expanding $ \left( x-1 \right)\left( x-b \right) $
We know that $ \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab $
Hence, we have
$ \left( x-1 \right)\left( x-b \right)={{x}^{2}}-\left( b+1 \right)x+b $
We know that that the discriminant of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given by $ D={{b}^{2}}-4ac $
Here a = 1, b = -(b+1) and c = b
Hence the discriminant of the expression is $ {{D}_{2}}={{\left( b+1 \right)}^{2}}-4b $
Using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ , we get
$ {{D}_{1}}={{b}^{2}}+1+2b-4b={{b}^{2}}-2b+1 $
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $
Hence, we have
$ {{D}_{1}}={{\left( b-1 \right)}^{2}} $
Since the discriminants of the expressions are equal, we have
$ \begin{align}
& {{\left( a-1 \right)}^{2}}={{\left( b-1 \right)}^{2}} \\
& \Rightarrow a-1=\pm \left( b-1 \right) \\
\end{align} $
Taking the positive sign, we get
$ a-1=b-1\Rightarrow a=b $
Taking the negative sign, we get
$ \begin{align}
& a-1=1-b \\
& \Rightarrow a+b=2 \\
\end{align} $
Hence the roots are either equal, or their sum is 2.
Note: The most common mistake done by the students is that they choose the equations to be $ {{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0 $ and $ {{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0 $ . This dramatically increases the difficulty of the question as the number of variables required to solve the question has been increased from 2 to 6. This practice should be avoided, and we often should think of a method that uses fewer variables as it makes the question easier to solve and also prevents us from making calculation mistakes to a great degree.
Complete step-by-step answer:
Let the other root of the first expression be a and the other root of the second expression be b
Since both the equations have x =1 as one of their roots, we have the quadratic expressions are
$ \left( x-1 \right)\left( x-a \right)=0\text{ and }\left( x-1 \right)\left( x-b \right)=0 $
Expanding $ \left( x-1 \right)\left( x-a \right) $
We know that $ \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab $
Hence, we have
$ \left( x-1 \right)\left( x-a \right)={{x}^{2}}-\left( a+1 \right)x+a $
We know that that the discriminant of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given by $ D={{b}^{2}}-4ac $
Here a = 1, b = -(a+1) and c = a
Hence the discriminant of the expression is $ {{D}_{1}}={{\left( a+1 \right)}^{2}}-4a $
Using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ , we get
$ {{D}_{1}}={{a}^{2}}+1+2a-4a={{a}^{2}}-2a+1 $
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $
Hence, we have
$ {{D}_{1}}={{\left( a-1 \right)}^{2}} $
Expanding $ \left( x-1 \right)\left( x-b \right) $
We know that $ \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab $
Hence, we have
$ \left( x-1 \right)\left( x-b \right)={{x}^{2}}-\left( b+1 \right)x+b $
We know that that the discriminant of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ is given by $ D={{b}^{2}}-4ac $
Here a = 1, b = -(b+1) and c = b
Hence the discriminant of the expression is $ {{D}_{2}}={{\left( b+1 \right)}^{2}}-4b $
Using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ , we get
$ {{D}_{1}}={{b}^{2}}+1+2b-4b={{b}^{2}}-2b+1 $
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $
Hence, we have
$ {{D}_{1}}={{\left( b-1 \right)}^{2}} $
Since the discriminants of the expressions are equal, we have
$ \begin{align}
& {{\left( a-1 \right)}^{2}}={{\left( b-1 \right)}^{2}} \\
& \Rightarrow a-1=\pm \left( b-1 \right) \\
\end{align} $
Taking the positive sign, we get
$ a-1=b-1\Rightarrow a=b $
Taking the negative sign, we get
$ \begin{align}
& a-1=1-b \\
& \Rightarrow a+b=2 \\
\end{align} $
Hence the roots are either equal, or their sum is 2.
Note: The most common mistake done by the students is that they choose the equations to be $ {{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0 $ and $ {{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0 $ . This dramatically increases the difficulty of the question as the number of variables required to solve the question has been increased from 2 to 6. This practice should be avoided, and we often should think of a method that uses fewer variables as it makes the question easier to solve and also prevents us from making calculation mistakes to a great degree.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE