NTSE Stage 2 SAT 2017 Question Paper

Let be the two quadratic equation since 1 is a common root.
Then,
⇒ ${\mathrm{a}}_1+{\mathrm{b}}_1+{\mathrm{c}}_1={\mathrm{a}}_2+{\mathrm{b}}_2+{\mathrm{c}}_2=0$
⇒ ${\mathrm{b}}_1=-\left({\mathrm{a}}_1+{\mathrm{c}}_1\right)\mathrm{and}{\mathrm{b}}_2=-\left({\mathrm{a}}_2+{\mathrm{c}}_2\right)$
Also, the discriminants of two quadratic equations are equal
Then
⇒ ${\mathrm{b}}_1^2-4{\mathrm{a}}_1{\mathrm{c}}_1={\mathrm{b}}_2^2-4{\mathrm{a}}_2{\mathrm{c}}_2$
⇒ ${\left({\mathrm{a}}_1-{\mathrm{c}}_1\right)}^2={\left({\mathrm{a}}_2-{\mathrm{c}}_2\right)}^2$
⇒ ${\mathrm{a}}_1-{\mathrm{c}}_1=\pm \left({\mathrm{a}}_2-{\mathrm{c}}_2\right)$
Now, the roots of equation,
$\mathrm{x}=-\frac{{\mathrm{b}}_1\pm \sqrt{{\mathrm{D}}_1}}{2}={\mathrm{a}}_1\mathrm{and}{\mathrm{c}}_1$ and the roots of equation $2,\mathrm{x}=\frac{-{\mathrm{b}}_2\pm \sqrt{{\mathrm{\Delta }}_2}}{2}={\mathrm{a}}_1$and ${\mathrm{c}}_2$
If ${\mathrm{a}}_1={\mathrm{a}}_2=1$ be the common roots
Then, other roots are ${\mathrm{c}}_1\mathrm{and}{\mathrm{c}}_2$
$\Rightarrow {\mathrm{c}}_1-{\mathrm{c}}_2={\mathrm{a}}_1-{\mathrm{a}}_2=0\mathrm{or}{\mathrm{c}}_1+{\mathrm{c}}_2={\mathrm{a}}_1+{\mathrm{a}}_2$
⇒ ${\mathrm{c}}_1={\mathrm{c}}_2{\mathrm{c}}_1+{\mathrm{c}}_2=2$
Hence, roots are either equal or their sum is 2.