UPSC CDS 2 2021 Math Paper

Let roots of the quadratic equation be a and b. Then,
a+b‌‌=7. . . (i)
a⋅b‌‌=12. . . (ii)
Put b=7−a into Eq. (ii), we get
a(7−a)‌=12
⇒‌‌‌‌7a−a2‌=12
⇒‌‌a2−7a+12‌=0
⇒a2−3a−4a+12‌=0
⇒a(a−3)−4(a−3)‌=0
⇒(a−3)(a−4)=0‌
a=3‌ or ‌4‌
‌ When, ‌a=3,b=7−3=4
‌ When, ‌a=4,b=7−4=3‌
So, the smaller root is 3 and the bigger root is 4 .
According to the question,
New roots are ‌

1

2

×(4) and 2×(3) i.e., 2 and 6.
The resulting quadratic equation :
x2−( sum of roots )x+ (product of roots )=0
⇒x2−(2+6)x+2×6=0
⇒x2−8x+12=0
Hence, the required quadratic equation is x2−8x+12=0