Advertisements
Advertisements
Question
`x^2-(sqrt3+1)x+sqrt3=0`
Solution
`x^2-(sqrt3+1)x+sqrt3=0`
⇒`x^2-sqrt3x-x+sqrt3=0`
⇒`x(x-sqrt3)-1 (x-sqrt3)=0`
⇒`(x-sqrt3)(x-1)=0`
⇒`x-sqrt3 or x-1=0`
⇒` x= sqrt3 or x=1`
Hence, 1 and `sqrt3` are the roots of the given equation.
APPEARS IN
RELATED QUESTIONS
Solve the equation 4x2 – 5x – 3 = 0 and give your answer correct to two decimal places
2x2 – 9x + 10 = 0, When
(1) x∈ N
(2) x∈ Q
Find the quadratic equation, whose solution set is: {3,5}
Solve the following equation for x and give, in the following case, your answer correct to 2 decimal places:
x2 – 5x – 10 = 0
Use the substitution y = 2x + 3 to solve for x, if` 4 (2x + 3)^2 − (2x + 3) − 14 = 0 `
`6x^2+x--12=0`
One root of the quadratic equation 8x2 + mx + 15 = 0 is `3/4`. Find the value of m. Also, find the other root of the equation.
Solve `9("x"^2 + 1/"x"^2) -3("x" - 1/"x") - 20 = 0`
Which of the following are quadratic equation:
(2x - 3) (x + 5) = 2 - 3x
Solve the following equation by using formula :
a (x2 + 1) = (a2+ 1) x , a ≠ 0
