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The sum of a number and its square is 63/4. Find the numbers.
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Let the number be x.
Given that the sum of x and its square = 63/4
⇒ x + ЁЭСе2 = 63/4
⇒ 4x + 4x2 - 63 = 0
⇒ 4x2 + 4x - 63 = 0
⇒ 4x2 + 4x - 63 = 0 ....................(i)
The value of x can be found by the formula
`x=(-b+-sqrt(b^2-4ac))/(2a)`
⇒ here a = 4, b = 4 and c = -63 from (i)
`x=(-4+-sqrt(16-4xx4xx-63))/(2xx4)`
`=(-4+-sqrt(16+16xx63))/(2xx4)`
`x=(-4+-sqrt(16+1008))/8`
Therefore,
`x=(-4+sqrt(16+1008))/8=7/2`
Or
`x=(-4-sqrt(16+1008))/8=-9/2`
∴ The values of x i. e. , the numbers is 7/2 , -9/2.
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