If a, b, c, d are real numbers such that a

Question

(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and d

Vedclass · Accepted Answer

Real and distinct

Vedclass · Answer

(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and distinct if the discriminant $D > 0$.Alternatively,consider the values of $f(x)$ at $x=a, b, c, d$:$f(a) = (a-a)(a-c) + 2(a-b)(a-d) = 2(a-b)(a-d)$. Since $a  0$.$f(b) = (b-a)(b-c) + 2(b-b)(b-d) = (b-a)(b-c)$. Since $b > a$ and $b  0$ and $(b-c) $f(c) = (c-a)(c-c) + 2(c-b)(c-d) = 2(c-b)(c-d)$. Since $c > b$ and $c  0$ and $(c-d) $f(d) = (d-a)(d-c) + 2(d-b)(d-d) = (d-a)(d-c)$. Since $d > a$ and $d > c$,$(d-a) > 0$ and $(d-c) > 0$,so $f(d) > 0$.Since $f(a) > 0$ and $f(b) Since $f(c)  0$,there exists a root in $(c, d)$.Thus,the equation has two real and distinct roots.

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

Similar Questions

Solve the equation $x^{2}-2x+\frac{3}{2}=0$.

The integral values of $a$ for which the quadratic equation $(x - a)(x - 10) + 1 = 0$ has integral roots are

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the roots of the equation $ax^2 - 2bx + c = 0$ are

If $\alpha$ is a root of multiplicity $3$ of the equation $x^5-8x^4+25x^3-38x^2+28x-8=0$,then $\alpha^2-5\alpha+6=$

The equation $x^2-4x+[x]+3=x[x]$,where $[x]$ denotes the greatest integer function,has:

Vedclass Test Series

Exam Paper Generator

Online Exam Module