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[a] Are either equal or their sum is 2

[b] Have to be always equal

[c] Are either equal or their sum is 1

[d] Have their sum to be equal to 1.

Answer
Verified

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.

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.

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