Question

Hii everyone!!! plz solve these questions

Answer 1

hope this helps you☺️

Answer 2

Hi there,

Here's the answers of 3,4,5

Ans3;

Let one of the odd positive integer be x

then the other odd positive integer is x+2

their sum of squares = x² +(x+2)²

= x² + x² + 4x +4

= 2x² + 4x + 4

Given that their sum of squares = 290

⇒ 2x² +4x + 4 = 290

⇒ 2x² +4x = 290-4 = 286

⇒ 2x² + 4x -286 = 0

⇒ 2(x² + 2x - 143) = 0

⇒ x² + 2x - 143 = 0

⇒ x² + 13x - 11x -143 = 0

⇒ x(x+13) - 11(x+13) = 0

⇒ (x-11) = 0 , (x+13) = 0

Therfore , x = 11 or -13

We always take positive value of x

So , x = 11

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Ans 4;

Let the smaller natural number be x and larger natural number be y

Hence x2 = 4y → (1)

Given y2 – x2 = 45

⇒ y2 – 4y = 45

⇒ y2 – 4y – 45 = 0

⇒ y2 – 9y + 5y – 45 = 0

⇒ y(y – 9) + 5(y – 9) = 0

⇒ (y – 9)(y + 5) = 0

⇒ (y – 9)= 0 or (y + 5) = 0

∴ y = 9 or y = -5

But y is natural number, hence y ≠ - 5

Therefore, y = 9

Equation (1) becomes,

x2 = 4(9) = 36

∴ x = 6

Thus the two natural numbers are 6 and 9.

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Ans 5;

Let the smaller part be x

Hence the larger part is (16 – x)

Given 2(16 – x)2 = x2 + 164

⇒ 2( 256 – 32x + x2) = x2 + 164

⇒ 512 – 64x + 2x2 – x2 – 164 = 0

⇒ x2– 64x + 348 = 0

=) (X-58) (X-6)

X= 58 OR X = 6

So, x = 58 is not possible

hence,smaller x= 6

and larger = 16-x = 16-6= 10.

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Thank you,

hope this help you :))