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प्रश्न
`2/x^2-5/x+2=0`
उत्तर
`2/x^2-5/x+2=0`
`(2-5x+2x^2)/x^2=0`
`2x^2-5x+2=0`
`4x^2-10x+4=0` (Multiplying both sides by 2)
`4x^2-10x=-4`
`(2x)^2-2xx2x xx5/2+(5/2)^2=-4+(5/2)^2` [Adding `(5/2)^2` on both sides ]
`(2x-5/2)^2=-4+25/4=9/4=(3/2)^2`
`2x-5/2=+-3/2` (Taking square root on both sides)
`2x-5/2=3/2 or 2x-5/2=-3/2`
`2x=3/2+5/2=8/2=4` or `2x=-3/2+5/2=2/2=1`
`x=2 or x=1/2`
Hence, 2 and `1/2` are the roots of the given equation.
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संबंधित प्रश्न
Solve the following quadratic equation by completing square method : x2 + 10x + 21 = 0.
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`4x^2+4bx-(a^2-b^2)=0`
The sum of the areas of two squares is `640m^2` . If the difference in their perimeter be 64m, find the sides of the two square
Find the value of discriminant.
2y2 – 5y + 10 = 0
Find the value of discriminant.
`sqrt2x^2 + 4x + 2sqrt2 = 0`
The sum of the squares of two consecutive natural numbers is 313. The numbers are:
If a, b, care in continued proportion, then show that `(a + b)^2/(ab) = (b + c)^2/(bc)`.
Since a, b, c are in continued proportion
∴ `a/b = square/square` = k(say)
⇒ b = `square`, a = `square` = `square`.k = `square`.k2
Now, L.H.S. = `(a + b)^2/(a.square) = (square + square)^2/(square*square)`
= `(squarek^2(k + 1)^2)/(square*square)`
= `(k + 1)^2/square`
R.H.S. = `(b + c)^2/(b.square) = (square + square)^2/(square*square) = (square (k + 1)^2)/(square*square)`
= `(k + 1)^2/square`
L.H.S. = `square`
