Advertisements
Advertisements
Question
`x^2-(1+sqrt2)x+sqrt2=0`
Solution
`x^2-(1+sqrt2)x+sqrt2=0`
⇒` x^2-x-sqrt2x+sqrt2=0`
⇒`x(x-1)-sqrt2(x-1)=0`
⇒` (x-sqrt2)(x-1)=0`
⇒`x=sqrt2=0 or x-1=0 `
⇒`x=sqrt2 or x=1`
Hence, the roots of the equation are `sqrt2` and 1
APPEARS IN
RELATED QUESTIONS
Write the following quadratic equation in a standard form: 3x2 =10x + 7.
In the following, determine whether the given values are solutions of the given equation or not:
`x+1/x=13/6`, `x=5/6`, `x=4/3`
In the following, determine whether the given values are solutions of the given equation or not:
a2x2 - 3abx + 2b2 = 0, `x=a/b`, `x=b/a`
Which of the following are quadratic equation in x?
`sqrt2x^2+7x+5sqrt2`
`x^2+6x+5=0`
`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3,x≠2,4`
The sum of 2 numbers is 18. If the sum of their reciprocals is `1/4` , find the numbers.
Solve:
`sqrt(x/(x - 3)) + sqrt((x - 3)/x) = 5/2`
Find whether the value x = `(1)/(a^2)` and x = `(1)/(b^2)` are the solution of the equation:
a2b2x2 - (a2 + b2) x + 1 = 0, a ≠ 0, b ≠ 0.
Solve the following equation by using formula :
2x2 – 3x – 1 = 0