Question

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

Answer 1

hey frnd....
let the two numbers be 'x' and 'y'
now according to question
x+y = 16
and 2x^2 = 164+ y^2

now by solving both the equation we get...our solution.

Hope this helps you ☺☺

Answer 2

Answer:

$\begin{gathered}\begin{gathered}\bf\: \sf \: \begin{cases} &\sf{Larger \: part \: = \: 10} \\ \\ &\sf{Smaller \: part \: = \: 6} \end{cases}\end{gathered}\end{gathered}$

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\bf\: Let\:assume\:that-\begin{cases} &\sf{larger \: part \: be \: x} \\ \\ &\sf{smaller \: part \: be \: 16 - x} \end{cases}\end{gathered}\end{gathered}$

According to statement, twice the square of the larger part exceeds the square of the smaller part by 164.

$\sf \: {2x}^{2} - {(16 - x)}^{2} = 164$

$\sf \: {2x}^{2} - [ {(16)}^{2} + {x}^{2} - 2(16)(x)] = 164$

$\sf \: {2x}^{2} - (256 + {x}^{2} - 32x) = 164$

$\sf \: {2x}^{2} -256 - {x}^{2} + 32x = 164$

$\sf \: {x}^{2} -256 + 32x = 164$

$\sf \: {x}^{2} + 32x - 256 - 164 = 0$

$\sf \: {x}^{2} + 32x - 420 = 0$

$\sf \: {x}^{2} + 42x - 10 - 420 = 0$

$\sf \: x(x + 42) - 10(x + 42) = 0$

$\sf \: (x + 42)(x - 10) = 0$

$\implies\sf \: x = - 42 \: \{rejected \} \: \: or \: \: x = 10$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \implies\sf \: \begin{cases} &\sf{Larger \: part \: = \: 10} \\ \\ &\sf{Smaller \: part \: = \: 6} \end{cases}\end{gathered}\end{gathered}$

$\rule{190pt}{2pt}$

Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac