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Case 1

If α and β are the roots of the equation ax2 + bx + c = 0, then we can write ax2 + bx + c = x2 – (α + β)x + α β = x2 – (sum of the roots) x + product of the roots = 0
Or, ax2 + bx + c = a(x − α)(x − β) = 0
 
In general, if x1x2x3,…, xn are the roots of the equation f(x) = axn + bxn-1 + cxn-2 +…+ K = 0
Then f(x) = axn + bxn-1 + cxn-2 +…+ K = a(x − x1) (x – x2) (x – x3)…(x – xn)

Case 2

p and q are the roots of the equation ax2 + bx + c = 0, and the quadratic equation with (1/p) and (1/q) as the roots is to be found:
 
Put x = 1/p
 
a(1/p)2 + b(1/p) + c = 0
 
or, cp2 + bp + a = 0
 
So, the quadratic equation is: cx2 + bx + a = 0

Case 3

p and q are the roots of the equation ax2 + bx + c = 0, and the quadratic equation with (−p) and (−q) as the roots is to be found:
 
Put x = − p
 
a(−p)2 + b(−p) + c = 0
 
Or, cp2 − bp + a = 0
 
So, the quadratic equation is: cx2 − bx + a = 0
 
Example
What will be the quadratic equation having the roots opposite in sign as that of the quadratic equation x2 + 5x + 6 = 0?
Solution
Putting (–x) at the place of x will give us the solution.
 
Desired equation is (−x)2 + 5 (−x) + 6 = 0
 
So, equation is: x2 − 5x + 6 = 0
 
Alternatively, the roots of the quadratic equation x2 + 5x + 6 = 0 are –2 and –3.
 
So, the roots of the desired equation should be 2 and 3.
 
So, the quadratic equation is (x−2) (x−3) = 0, or, x2 −5x + 6 = 0
 

Case 4

p and q are the roots of the equation ax2 + bx + c = 0, and the quadratic equation with (p)2 and (q)2 as the roots is to be found:
 
Let x be a root of the original equation and y be a root of the transformed equation.
 
y = x2, so we will put x = √y
 
a(√y)2 + b(√y) + c = 0,
 
Solving this will give us the desired equation.





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