Advertisements
Advertisements
प्रश्न
`5x^2-6x-2=0`
उत्तर
`5x^2-6x-2=0`
⇒`25x^2-30x-10=0`
⇒`25x^2-30x=10`
⇒`(5x)^2-xx5x xx3+3^2=10+3^3` (Adding `3^2` on both sides)
⇒`(5x-3)^2=10+9-19`
⇒`5x-3=+-sqrt19` (Taking square root on both)
⇒`5x-3=sqrt19 or 5x-3=-sqrt19`
⇒`5x=3+-sqrt19 or 5x=3-sqrt19`
⇒`x=(3+sqrt19)/5 or x=(3-sqrt19)/5`
Hence, `(3+sqrt19)/5` and `(3-sqrt19)/5` are `(3-sqrt19)/5` are the roots of the given equation.
APPEARS IN
संबंधित प्रश्न
Find the roots of the following quadratic equations, if they exist, by the method of completing the square 2x2 + x – 4 = 0
The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.
`4x^2+4bx-(a^2-b^2)=0`
`x^2-(sqrt2+1)x+sqrt2=0`
Solve the following quadratic equation for x:
`4sqrt3x^2+5x-2sqrt3=0`
Determine the nature of roots of the following quadratic equation.
m2 + 2m + 9 = 0
Form the quadratic equation from the roots given below.
3 and –10
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages 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`
