Question

if one root of the equation ax²+bx+c=0 is the cube of the other, prove that,(b²-2ac)²=ca(c+a)²​

Answer 1

Step-by-step explanation:

How will I prove (b^2-2ac) ^2=CA(c+a) ^2 when one root of the quadratic equation ax^2+bx+c=0 is the cube of the other root?

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Let the 2 roots of the quadratic equation ax2+bx+c=0 be α,α3

Sum of the roots: α+α3=−ba—(1)

Product of the roots: α.α3=α4=ca

⟹α=(ca)14—(2)

Squaring (1):

(α+α3)2=(−ba)2

⟹α2+α6+2α4=b2a2

⟹α2(1+α4)+2α4=b2a2−−(3)

Substituting (2) in (3):

⟹α2(1+ca)+2ca=b2a2

⟹α2.a+ca=b2a2−2ca=b2−2aca2

⟹α2=b2−2aca(a+c)−−(4)

Squaring (4):

α4=(b2−2aca(a+c))2=(b2−2ac)2a2(a+c)2−−(5)

Substituting (2) in (5):

ca=(b2−2ac)2a2(a+c)2

⟹ac(a+c)2=(b2−2ac)2