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Algebra Level pending

In a isosceles triangle △ A B C \bigtriangleup ABC , a , b , c a,b,c are the sides.

Out of which two sides ( a , b ) (a,b) are equal to two distinct roots of equation:-

16 x 3 − 124 x 2 + 280 x − 147 = 0 16x^3-124x^2+280x-147=0 .

The other side ( c ) (c) is equal to either a a or b b that depends on the given condition:- a + c = a+c= integer

Find a + b + c = ? a+b+c=?


The answer is 7.75.

Akash Shukla by Akash Shukla , 26, India

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1 solution

Tom Engelsman
Aug 23, 2017

The above cubic equation is expressible as:

16 x 3 − 124 x 2 + 280 x − 147 = 0 ⇒ ( 4 x − 3 ) ( 2 x − 7 ) 2 = 0 ⇒ x = 3 4 , 7 2 . 16x^3 - 124x^2 + 280x - 147 = 0 \Rightarrow (4x-3)(2x-7)^2 = 0 \Rightarrow x = \frac{3}{4}, \frac{7}{2}.

Hence we require ( a , b , c ) = ( 7 2 , 3 4 , 7 2 ) (a,b,c) = (\frac{7}{2}, \frac{3}{4}, \frac{7}{2}) , which leads to the sum a + b + c = 7.75 . a+b+c = \boxed{7.75}.

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