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Home » 6. Divide 25 into two parts such that twice the square of the larger part exceeds thrice the square of the smaller part by 29.
6. Divide 25 into two parts such that twice the square of the larger part exceeds thrice the square of the smaller part by 29.
CBSE | Class 10 | Class 10 | Frank | Maths | Problems Based On Quadratic Equations
6. Divide 25 into two parts such that twice the square of the larger part exceeds thrice the square of the smaller part by 29.

Solution:-

Let us assume the two numbers be A and B, B being the bigger number.

As per the condition given in the question,

A + B = 25 … [equation (i)]

    \[2{{B}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\]

… [equation (ii)]

Now, consider the equation (i),A + B = 25

B = 25 - A

Then, substitute the value of B is equation (ii),

    \[2{{\left( 25-A \right)}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\]

We know that,

    \[{{\left( a-b \right)}^{2}}~=\text{ }{{a}^{2}}~-2ab\text{ }+\text{ }{{b}^{2}}\]

    \[2({{25}^{2}}-\left( 2\text{ }\times \text{ }25\text{ }\times \text{ }A \right)\text{ }+\text{ }{{A}^{2}})\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]

    \[1250\text{ }+\text{ }2{{A}^{2}}-100A\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]

    \[{{A}^{2}}~+\text{ }100A-1221\text{ }=\text{ }0\]

    \[{{A}^{2}}~-11A\text{ }+\text{ }111A-1221\text{ }=\text{ }0\]

Take out common in each terms,

A(A - 11) + 111(A - 11) = 0

(A - 11) (A + 111) = 0

Equate both to zero,

A - 11 = 0

A + 111 = 0

A = 11

A = -111

Therefore, A = 11 … [because A can’t be negative]

So, B = 25 - A

= 25 - 11

= 14

i

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