\displaystyle \textbf{Question 1. }\text{Write the number of real roots of the equation } \\ (x-1)^2+(x-2)^2+(x-3)^2=0.
\displaystyle \text{Answer:}
\displaystyle (x-1)^2\geq0,\qquad (x-2)^2\geq0,\qquad (x-3)^2\geq0
\displaystyle \text{Sum of three non-negative quantities can be zero only if each is zero.}
\displaystyle \text{But }(x-1)^2,\ (x-2)^2,\ (x-3)^2\text{ cannot all be zero simultaneously.}
\displaystyle \therefore \text{No real root exists.}
\displaystyle \therefore \text{Number of real roots }=0
\\

\displaystyle \textbf{Question 2. }\text{If }a\text{ and }b\text{ are roots of the equation }x^2-px+q=0,\text{ then write the value of }\frac{1}{a}+\frac{1}{b}.
\displaystyle \text{Answer:}
\displaystyle \text{For the equation }x^2-px+q=0,
\displaystyle a+b=p\qquad \text{and}\qquad ab=q
\displaystyle \therefore \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}
\displaystyle =\frac{p}{q}
\\

\displaystyle \textbf{Question 3. }\text{If roots }\alpha,\beta\text{ of the equation }x^2-px+16=0\text{ satisfy } \\ \alpha^2+\beta^2=9,\text{ then write the value of }p.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=p,\qquad \alpha\beta=16
\displaystyle \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta
\displaystyle 9=p^2-32
\displaystyle \therefore p^2=41
\displaystyle \therefore p=\pm\sqrt{41}
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\displaystyle \textbf{Question 4. }\text{If }2+\sqrt3\text{ is a root of the equation }x^2+px+q=0,\text{ then write the values of } \\ p\text{ and }q.
\displaystyle \text{Answer:}
\displaystyle \text{Since the equation has rational coefficients, the other root is }2-\sqrt3
\displaystyle \therefore \text{Sum of roots }=(2+\sqrt3)+(2-\sqrt3)=4
\displaystyle \text{Product of roots }=(2+\sqrt3)(2-\sqrt3)=1
\displaystyle \therefore -p=4,\qquad q=1
\displaystyle \therefore p=-4,\qquad q=1
\\

\displaystyle \textbf{Question 5. }\text{If the difference between the roots of the equation } \\ x^2+ax+8=0\text{ is }2,\text{ write the values of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots be }\alpha\text{ and }\beta.
\displaystyle \alpha+\beta=-a,\qquad \alpha\beta=8,\qquad \alpha-\beta=2
\displaystyle (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta
\displaystyle 4=a^2-32
\displaystyle \therefore a^2=36
\displaystyle \therefore a=\pm6
\\

\displaystyle \textbf{Question 6. }\text{Write the roots of the equation }(a-b)x^2+(b-c)x+(c-a)=0.
\displaystyle \text{Answer:}
\displaystyle (a-b)x^2+(b-c)x+(c-a)=0
\displaystyle \text{Putting }x=1,\text{ we get }(a-b)+(b-c)+(c-a)=0
\displaystyle \therefore x=1\text{ is one root.}
\displaystyle \text{Product of roots }=\frac{c-a}{a-b}
\displaystyle \therefore \text{Other root }=\frac{c-a}{a-b}
\displaystyle \therefore \text{Roots are }1,\ \frac{c-a}{a-b}
\\

\displaystyle \textbf{Question 7. }\text{If }a\text{ and }b\text{ are roots of the equation } \\ x^2-x+1=0,\text{ then write the value of }a^2+b^2.
\displaystyle \text{Answer:}
\displaystyle a+b=1,\qquad ab=1
\displaystyle a^2+b^2=(a+b)^2-2ab
\displaystyle =1^2-2\cdot1=-1
\\

\displaystyle \textbf{Question 8. }\text{Write the number of quadratic equations, with real roots, which do} \\ \text{not change by squaring their roots.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots be }\alpha,\beta.
\displaystyle \text{The roots remain unchanged after squaring, so }\{\alpha,\beta\}=\{\alpha^2,\beta^2\}
\displaystyle \therefore \text{Possible roots are }0\text{ and }1,\text{ or }\alpha^2=\beta,\ \beta^2=\alpha
\displaystyle \text{The distinct quadratic equations are }x^2,\ x(x-1),\ (x-1)^2,\ x^2+x+1
\displaystyle \text{But }x^2+x+1=0\text{ has non-real roots.}
\displaystyle \therefore \text{Number of such quadratic equations }=3
\\

\displaystyle \textbf{Question 9. }\text{If }\alpha,\beta\text{ are roots of the equation }x^2+lx+m=0, \\ \text{ write an equation whose roots are }-\frac{1}{\alpha}\text{ and }-\frac{1}{\beta}.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=-l,\qquad \alpha\beta=m
\displaystyle \text{New sum }=-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha\beta}=\frac{l}{m}
\displaystyle \text{New product }=\frac{1}{\alpha\beta}=\frac{1}{m}
\displaystyle \therefore \text{Required equation is }x^2-\frac{l}{m}x+\frac{1}{m}=0
\displaystyle \therefore mx^2-lx+1=0
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\displaystyle \textbf{Question 10. }\text{If }\alpha,\beta\text{ are roots of the equation }x^2-a(x+1)-c=0, \\ \text{ then write the value of }(1+\alpha)(1+\beta).
\displaystyle \text{Answer:}
\displaystyle x^2-a(x+1)-c=0
\displaystyle \Rightarrow x^2-ax-(a+c)=0
\displaystyle \alpha+\beta=a,\qquad \alpha\beta=-(a+c)
\displaystyle (1+\alpha)(1+\beta)=1+\alpha+\beta+\alpha\beta
\displaystyle =1+a-(a+c)=1-c
\\

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