Home
Class 10
MATHS
If a, b and c are distinct real numbers,...

If a, b and c are distinct real numbers, prove that the equation `(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0` has real and distinct roots.

Text Solution

Verified by Experts

The correct Answer is:
`D gt 0`
`:.` hence it has real and distinct roots.
Promotional Banner

Topper's Solved these Questions

  • QUADRATIC EQUATIONS

    ZEN PUBLICATION|Exercise ZEN ADDITIONAL QUESTIONS (IIT FOUNDATION)|5 Videos

  • QUADRATIC EQUATIONS

    ZEN PUBLICATION|Exercise ZEN ADDITIONAL QUESTIONS (LONG ANSWER [LA] TYPE QUESTIONS)|17 Videos

  • PROBABILITY

    ZEN PUBLICATION|Exercise ZEN ADDITIONAL QUESTIONS HOTS [HIGHER ORDER THINKING SKILLS]-QUESTIONS|24 Videos

  • REAL NUMBERS

    ZEN PUBLICATION|Exercise ADDITIONAL QUESTIONS(LONG ANSWER TYPE QUESTIONS )|5 Videos

Similar Questions

Explore conceptually related problems

If a le b le c le d then the equation (x-a)(x-c)+2(x-b)(x-d)=0 has

If the roots of the equation (a-b)x^(2)+(b-c)x+(c-a)=0 are equal, then

A root of the equation |(0, x-a,x-b),(x+a,0,x-c),(x+b,x+c,0)|=0 is

If a, b, c are positive real numbers, then the number of real roots of the equation a x^(2)+b|x|+c=0 is

If a, b, c are real and a!=b , then the roots of the equation, 2(a-b)x^2-11(a + b + c) x-3(a-b) = 0 are :

Both the roots of the equation : (x - b) (x -c) + (x - c) (x - a) + (x - a) (x - b) = 0 are always :

Show that the equation 2(a^(2)+b^(2))x^(2)+2(a+b)x+1=0 has no real roots when a!=b .

If a ,b ,c ,d are four consecutive terms of an increasing A.P., then the roots of the equation (x-a)(x-c)+2(x-b)(x-d)=0 are a. non-real complex b. real and equal c. integers d. real and distinct

x and b are real numbers. If b gt 0 and |x| gt b , then