AP EAMCET20226 Jul 2022Evening ShiftMathematicsQuadratic EquationActual
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
Options
- AReal and need not be distinct
- BReal and distinct
- CNon-real and distinct
- DNon-real and need not be distinct
Correct answer
B. Real and distinct
Step-by-step solution
aligned & Here, (x-a)(x-c)+2(x-b)(x-d) & =x^2-(a+c) x+a c+2 [x^2-(b+d) x+b d ] & =x^2-(a+c) x+a c+2 x^2-2(b+d) x+2 b d & =3 x^2-(a+2 b+c+2 d)+a c+2 b d & D=b^2-4 a c & =(a+2 b+c+2 d)^2-12(a c+2 b d) & =[(a+2 d)+(2 b+c)]^2-12(a c+2 b d) & =[(a+2 d)-(2 b+c)]^2+4(a+2 d)(2 b+c) . -12(a c+2 b d) & =[(a+2 d)-(2 b+c)]^2+4 a c+8 a b+16 b d & =[(a+2 d)-(2 b+c)]^2-8 a c-8 b d+8 a b+8 c d & =[(a+2 d)-(2 b+c)]^2+8(c-b)(d-a) aligned As, a 0 Roots are real and distinct.