the roots of the equation a(b-c)x2 +b(c-a)x +c(a-b) =0 are
Answers
Given quadratic equation,
a(b-c)x^2+b(c-a)^2+c(a-b)=0
also zeroes of given equation are equal.
Therefore,
discriminant=0
b^2-4ac=0
{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0
b^2(c^2+a^2-2ac)-4{ab-ac}{ac-bc}=0
b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0
b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0
We know that,
a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2
By above identity we get,
{bc+ab-2ac}^2=0
bc+ab-2ac=0
b(a+c)=2ac
b=2ac/(a+c)
Hence if zeroes of given quadratic equation are equal then,
b=2ac/(a+c)
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hope it helps
Step-by-step explanation:
1
Secondary School Math 5 points
If the roots of the equation a(b-c)x2 +b(c-a)x +c(a-b) =0 are equal
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Answers
prajapatyk01
prajapatyk01 Ace
Given quadratic equation,
a(b-c)x^2+b(c-a)^2+c(a-b)=0
also zeroes of given equation are equal.
Therefore,
discriminant=0
b^2-4ac=0
{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0
b^2(c^2+a^2-2ac)-4{ab-ac}{ac-bc}=0
b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0
b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0
We know that,
a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2
By above identity we get,
{bc+ab-2ac}^2=0
bc+ab-2ac=0
b(a+c)=2ac
b=2ac/(a+c)
Hence if zeroes of given quadratic equation are equal then,
b=2ac/(a+c)