Math, asked by bhavuapbwb52, 1 year ago

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

Answers

Answered by ItzRadhika
3

Refers to attachment ~

Attachments:
Answered by mathdude500
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

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