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प्रश्न
`4x^2+4bx-(a^2-b^2)=0`
उत्तर
`4x^2+4bx-(a^2-b^2)=0`
`4x^2+4bx=a^2-b^2`
`(2x)^2+2xx2x xxb+b^2=a^2-b^2+b^2=a^2-b^2+b^2` (Adding `b^2` on both sides)
`(2x+b)^2=a^2`
`2x+b=+-a` (Taking square root on both sides)
`2x+b=a or 2x+b=-a`
`2x=a-b or 2x=-a-b`
`x=(a-b)/2 or x=-(a+b)/2`
Hence, `(a-b)/2` and `-(a+b)/2` are the roots of the given equation.
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संबंधित प्रश्न
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Complete the following activity to solve the given word problem. The Sum of squares of two consecutive even natural numbers is 244, then find those numbers.
Activity: Let the first even natural number be x
Therefore its consecutive even natural number will be = (______)
By the given condition,
x2 + (x + 2)2 = 244
x2 + x2 + 4x + 4 – (______) = 0
2x2 + 4x – 240 = 0
x2 + 2x – 120 = 0
x2 + (______) – (______) – 120 = 0
x(x + 12) – (______) (x + 12) = 0
(x + 12)(x – 10) = 0
x = (______)/x = 10
But natural number cannot be negative, x = – 12 is not possible.
Therefore first even natural number is x = 10.
Second even consecutive natural number = x + 2 = 10 + 2 = 12.
