Question

sum of the square of two consecutive positive even integers is 100 find two. number using quadratic equations ​

Answer 1

Answer:

6 and 8.

Step-by-step explanation:

sum of the square of two consecutive positive even integers = 100

Since they are even, Let the integers be x and x+2.

According to the question, x² + (x+2)² = 100

⇒ x² + x² + 2×x×2 + 2² = 100

⇒ x² + x² +4x + 4 = 100

⇒ 2x² + 4x + 4 = 100

⇒ 2x² + 4x + 4 - 100 = 0

⇒ 2x² + 4x - 96 = 0

⇒ x² + 2x - 48 = 0

You need to solve the quadratic equation by any method you like. I am using middle term splitting.

⇒ x² + 2x - 48 = 0

⇒ x² + 8x - 6x - 48 = 0

⇒ x(x+8) - 6(x+8) = 0

⇒ (x+8)(x-6) = 0

⇒ x+8 = 0   OR     x-6 =0

⇒ x = -8 (not accepted as x is positive)

OR x = 6

x+2 = 6+2 = 8

Hence, NUmbers are 6 and 8

Answer 2

Given:-

Sum of the square of two consecutive positive even integers is 100 find two. number using quadratic equations .

To find:-

The numbers.

Solution:-

• x²+(x+2)²=100
• Using (a+b)²=a²+2ab+b²
• x²+x²+2×x×2+2²=100
• 2x²+4x+4=100
• 2x²+4x+4-100=0
• 2x²+4x-96=0
• 2x²/2+4x/2-96/2=0
• x²+2x-48=0

Now we have a quadratic equation.

• x²+8x-6x-48=0
• Taking x and 8 common.
• x(x+8)-6(x+8)=0
• Either
• x+8=0
• or
• x-6=0
• On further calculation of x+8 we get
• x=-8
• On further calculation of x-6 we get
• x=6
• As x=-8 is not acceptable so the two numbers are
• 8 and 6

Henceforth,the two numbers whose square when added gives 100 are 8 and 6.

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