Question

help me with 25th sum...​

Answer 1

$\mathfrak{\huge{The\:Question\:Says:}}$

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

$\mathfrak{\huge{Answer:}}$

Let the parts be = x and y

We know that these numbers are parts of 20. Thus, we can also say that:

$\bf{x + y = 20}$ .....(1)

Now, the given condition is that $\bf{3\: times \: the \:square}$ of one part exceeds the other part $\bf{by\:10}$

=》 $\bf{3x^{2} = y +10}$

=》 $\bf{3x^{2} - 10 = y}$ ....(2)

Put (2) in (1) :

=》 $\bf{x + y = 20}$

=》 $\bf{x + 3x^{2} - 10 = 20}$

=》 $\bf{3x^{2} + x - 30 = 0}$

Solve this formed equation further by the splitting middle term method:

=》 $\bf{3x^{2} - 9x + 10x - 30}$

=》 $\bf{3x(x - 3) + 10(x -3)}$

=》 $\bf{(3x+10)(x-3)}$

=》 $\bf{x = \frac{(-10)}{3} , 3}$

Since the first value isn't acceptable, the only value of x can be 3

Therefore , $\tt{x = 3}$

=》 x + y = 20

=》 $\tt{y = 17}$

Thus, the numbers will be $\huge{\sf{3\:and\:17}}$

Answer 2

Solution :

Assuming the parts as P and Q.

➨ P + Q = 20 ...........(i)

According to the Question ,

➨ 3p² = q + 10

➨ 3p² - 10 = q ............(ii)

Now , putting (ii) in (i) :-

➨ p + q = 20

➨ p + 3p² - 10 = 20

➨ 3p² + p - 30 = 0

Now ,

Let's Solve the equation -

➨ 3p² - 9p + 10p - 30

➨ 3p ( p - 3 ) + 10 ( p - 3 )

➨ ( 3p + 10 ) ( p - 3 )

➨ p = (-10)/3 and 3

Here , appropriate value of p can only be 3.

So , P = 3

➨ p + q = 20

➨ y = 20 - p i.e. 20 - 3

➨ y = 17

Hence ,

The numbers are 3 and 17.