Quadratic Equation Solution5





In this page quadratic equation solution5 we are going to see solution of the word problems of the topic quadratic equation.

Question 9

Divide 29 into two parts so that sum of their squares is 425.

Solution:

Let "x" and "y" are the required two numbers

Divide 29 into two parts

So let us take it as sum of two numbers is

 x + y = 29

      y = 29 - x -----(1)

sum of their squares is 425.

 x² + y² = 425

 x² + (29 - x)² = 425

 x² + (29)² - 2 (x) (29) + x² = 425

 2 x² + 841 - 58 x - 425 = 0

 2 x² - 58 x + 416 = 0

Now we are going to divide the whole equation by 2, So that we will get

 x² - 29 x + 208 = 0

 x² - 13 x - 16 x + 208 = 0

 x (x - 13) - 16 (x - 13) = 0

(x - 13) (x - 16) = 0

 x - 13 = 0             x - 16 = 0

 x = 13                     x = 16

Therefore the required numbers are 13 and 16

Verification:

sum of two numbers is 29

 13 + 16 = 29

   29 = 29

sum of their squares is 425

 13² + 16² = 425

 169 + 256 = 425

    425 = 425


Question 10

Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

Solution:

Let "x" be the required number

whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

 x - 20 =69 (1/x)

 x - 20 =69/x

x (x - 20) = 69

 x² - 20 x = 69

 x² - 20 x - 69 = 0

 x² - 23 x  + 3 x - 69 = 0

x (x - 23) + 3 (x - 23) = 0

(x + 3) (x - 23) = 0

 x + 3 = 0            x - 23 = 0.

    x = -3                x = 23

Therefore the required number is 23.

Verification:

whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

23 - 20 = 69 (1/23)

 3 = 69/23

 3 = 3

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