Advertisements
Advertisements
Question
`x^2-2ax+(a^2-b^2)=0`
Solution
Given:
`x^2-2ax+(a^2-b^2)=0`
On comparing it with `Ax^2+Bx+C=0` we get
`A=1, B=-2a and C=(a^2-b^2)`
Discriminant D is given by:
`D=B^2-4AC`
=`(-2a)^2-4xx1xx(a^2-b^2)`
=`4a^2-4a^2+4b^2`
=`4b^2>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt(D))/(2a)=(-(-2a)+sqrt4b^2)/(2xx1)=(2a+2b)/2=(2(a+b))/2=(a+b)`
`β=(-b-sqrt(D))/(2a)=(-(-2a)-sqrt4b^2)/(2xx1)=(2a-2b)/2=(2(a-b))/2=(a-b)`
Hence, the roots of the equation are (a +b) and (a +b).
APPEARS IN
RELATED QUESTIONS
Write the discriminant of the following quadratic equations:
2x2 - 5x + 3 = 0
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3a2x2 + 8abx + 4b2 = 0, a ≠ 0
Solve for x
`x+1/x=3`, x ≠ 0
Which of the following are the roots of `3x^2+2x-1=0`
`-1/2`
`2x^2+x-4=0`
`x+1/x=3,x≠0`
`x^2-4ax-b^2+4a^2=0`
`3a^2x^2+8abx+4b^2=0`
Find the nature of roots of the following quadratic equations:
`12x^2-4sqrt15x+5=0`
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.
