Question

If the discriminants of two equations are equal and the equations have a common root 1, then the other roots are[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.

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.

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.