Advertisements
Advertisements
Question
`x^2-6x+4=0`
Solution
Given:
`x^2-6x+4=0`
On comparing it with `ax^2+bx+c=0`
a =1,b = -6 and c = 4
Discriminant D is given by:
`D=(b^2-4ac) `
=`(-6)^2-4xx1xx4`
=`36-16`
=`20>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt(D))/(2c)=(-(-6)+sqrt(20))/(2xx1)=(6+2sqrt(5))/2=(2(3+sqrt(3)))/2=(3+sqrt(5))`
`β=(-b-sqrt(D))/(2a)=(-(-6)-sqrt(20))/2=(6-2sqrt(5))/2=(2(3-sqrt(5)))/2=(3-sqrt(5))`
Thus, the roots of the equation are `(3-2sqrt(5)) and (3-2sqrt(5))`
APPEARS IN
RELATED QUESTIONS
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3x2 - 2x + 2 = 0
Solve for x
`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`; x ≠ 2, 4
`3x^2-243=0`
`sqrt3x^2-2sqrt2x-2sqrt3=0`
`x^2-2ax+(a^2-b^2)=0`
`x^2-4ax-b^2+4a^2=0`
`x^2-(2b-1)x+(b^2-b-20)=0`
If 3 is a root of the quadratic equation` x^2-x+k=0` find the value of p so that the roots of the equation `x^2+2kx+(k^2+2k+p)=0` are equal.
If the roots of the quadratic equation `(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0` are real and equal, show that either a=0 or `(a^3+b^3+c^3=3abc)`
A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digit interchange their places. Find the number.
