find roots by sridharacharya method....
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First we have to calculate the discriminant
2x^2 +ax+a^2
D = b^2-4ac
= a^2-4×2×(-a^2)
= a^2+8a^2
= 9a^2
Now by applying formula
X = b+_- √D/2×a
= a +_- √9a^2/2×2
= one time we will plus one time we will minus
a + 3a /4 = a
a-3a /4 = - a/2
Remember shri dharacharya is other name of quadratic formula
2x^2 +ax+a^2
D = b^2-4ac
= a^2-4×2×(-a^2)
= a^2+8a^2
= 9a^2
Now by applying formula
X = b+_- √D/2×a
= a +_- √9a^2/2×2
= one time we will plus one time we will minus
a + 3a /4 = a
a-3a /4 = - a/2
Remember shri dharacharya is other name of quadratic formula
prakharkumar89:
your answer is very helpful for me thanks
welcome
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