Question

A positive real number is 3 more than another. When -6 times the smaller is added to the square of the larger, the result is 59. Find the numbers.

Answer 1

Solution :-

Let the small number be x

The smaller number is 3 times the larger number

Therefore,

The larger number will be x + 3

Now,

According to the question,

When -6 is added to the square of the larger number then the result is 59

That is,

-6x + ( x + 3)² = 59

-6x + x² + 6x + 9 = 59

[ Using identity ( a + b)² = a² + b² + 2ab ]

x² + 9 = 59

x² = 59 - 9

x² = 50

x = √50

x = 5√2

The small number is = 5√2

The larger number is = 5√2 + 3 = 8√2

Hence, The two numbers are 5√2 and 8√2 $.$

Answer 2

Given:-

• A positive real number is 3 more than another.
• When -6 times the smaller is added to the square of the larger, the result is 59.

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To find:-

• The numbers.

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Solution:-

★ In this question we given that a positive real number is 3 more than another. When -6 times the smaller is added to the square of the larger, the result is 59. We have to find out the numbers. Let's do it.

⠀

Let,

• The small number be x.
• The larger number be x + 3

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According to the question,

⠀

⇢ -6x + ( x + 3)² = 59

⇢ -6x + x² + 6x + 9 = 59

⇢ x² + 9 = 59

⇢ x² = 59 - 9

⇢ x² = 50

⇢ x = √50

⇢ x = 5√2

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Therefore,

• Smaller number = 5√2
• Larger number = 5√2 + 3 = 8√2

⠀

Hence,

• The numbers are 5√2 and 8√2.