Question

If \(a\), \(b\), \(c\), \(d\) are real numbers such that \[ a\lt b\lt c\lt d, \] then the roots of the equation \[ (x-a)(x-c)+2(x-b)(x-d)=0 \] are

• A. Real & need not be distinct
• B. Real and distinct
• C. Non-real and distinct
• D. Non-real and need not be distinct

Hint:

For quadratic equations, always use the discriminant: \[ \Delta=b^2-4ac \] If \[ \Delta\gt 0, \] the roots are real and distinct.

Answer 1

The Correct Option is B

Step 1: Expand the given equation.
Given, \[ (x-a)(x-c)+2(x-b)(x-d)=0 \] Expanding, \[ x^2-(a+c)x+ac + 2(x^2-(b+d)x+bd)=0 \] \[ x^2-(a+c)x+ac+2x^2-2(b+d)x+2bd=0 \] \[ 3x^2-\left(a+c+2b+2d\right)x+(ac+2bd)=0 \] Thus, the quadratic equation is \[ 3x^2-\left(a+c+2b+2d\right)x+(ac+2bd)=0 \]

Step 2: Find the discriminant.
For a quadratic equation \[ Ax^2+Bx+C=0, \] the discriminant is \[ \Delta=B^2-4AC \] Here, \[ A=3 \] \[ B=-(a+c+2b+2d) \] \[ C=ac+2bd \] Therefore, \[ \Delta=(a+c+2b+2d)^2-12(ac+2bd) \]

Step 3: Simplify the discriminant.
Rewrite the discriminant as \[ \Delta=(a-c)^2+4(b-d)^2 \] \[ +4(a-b)(c-d)+4(a-d)(b-c) \] Since \[ a\lt b\lt c\lt d, \] we have \[ a-b\lt 0,\quad c-d\lt 0 \] so \[ (a-b)(c-d)\gt 0 \] Also, \[ a-d\lt 0,\quad b-c\lt 0 \] so \[ (a-d)(b-c)\gt 0 \] Further, \[ (a-c)^2\gt 0 \] and \[ (b-d)^2\gt 0 \] Hence, \[ \Delta\gt 0 \]

Step 4: Conclude about the roots.
Since \[ \Delta\gt 0, \] the quadratic equation has two real and distinct roots.

Step 5: Final conclusion.
Therefore, the roots are \[ \boxed{\text{Real and distinct}} \]