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 equation into standard quadratic form.
$(x-a)(x-c)+2(x-b)(x-d)=0$. Expanding: $x^2-(a+c)x+ac + 2x^2-2(b+d)x+2bd = 0$. Combining: $3x^2-(a+c+2b+2d)x+(ac+2bd)=0$.
Step 2: Evaluate $f(x) = (x-a)(x-c)+2(x-b)(x-d)$ at $x=b$.
At $x=b$: $f(b) = (b-a)(b-c)+2(0)(b-d) = (b-a)(b-c)$. Since $a < b < c$: $b-a > 0$ and $b-c < 0$, so $f(b) < 0$.
Step 3: Evaluate $f$ at $x=d$.
At $x=d$: $f(d) = (d-a)(d-c)+0 = (d-a)(d-c)$. Since $a < c < d$: $d-a > 0$ and $d-c > 0$, so $f(d) > 0$.
Step 4: Apply the Intermediate Value Theorem between $b$ and $d$.
Since $f$ is a polynomial (continuous), $f(b) < 0$ and $f(d) > 0$ imply there is a root in $(b,d)$. Also, as $x \to -\infty$, $f(x) \to +\infty$ (leading coefficient $3 > 0$), and $f(b) < 0$, so there is another root in $(-\infty, b)$.
Step 5: Confirm the two roots are distinct.
We located one root in $(-\infty, b)$ and another in $(b,d)$. These are disjoint intervals, so the two roots are distinct real numbers.
Step 6: State the final answer.
The equation has real and distinct roots. \[ \boxed{\text{Real and distinct}} \]