Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Find the two parts.

Open in App

Solution

Let One part be x.
So, the other part is $(20 - x)$
According to question.
$3 x^{2} = (20 - x) + 10$
$3 x^{2} = 30 - x$
$3 x^{2} + x - 30 = 0$
$3 x^{2} + 10 x - 9 x - 30 = 0$
$x (3 x + 10) - 3 (3 x + 10) = 0$
$(3 x + 10) (x - 3) = 0$
$x = 3$ , $x = \frac{- 10}{3}$
Hence, the two parts $= 3, 17$

Suggest Corrections

Similar questions

20

10

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

900

60 %

30 %

270

16

164.

184

8

Join BYJU'S Learning Program