Please Solve Q.1.. Thank You..

Please Solve Q.1.. Thank You.. Exercise-2 PART - I : ONLY ONE OPTION CORRECT TYPE Let a > O, b > O & c > 0. Then both the roots of the equation ax? + bx + c = 0 (A) are real & negative (C) are rational numbers (B) have negative real parts (D) have positive real parts 2. 3. If the roots of the equation x2 + 2ax + b = O are real and distinct and they differ by at most 2m, then b lies in the interval (A) (a2 — rn2, a2) — m2, a?) (C) (a2, az + m2) (D) none of these The set of possible values of R for which x2 — — 5k + 5)x + (22,2 — 3).—4) = O has roots, whose sum and product are both less than 1, is 5 2 5 2 5 2

Consider equation ax2+bx+c=0Discriminant of the equation, D=b2-4acb2>0 and 4ac>0 as a,b,c>0Therefore, we cannot comment on sign if b2-4acHence, we cannot say if roots are real or imaginary.ax2+bx+c=0x=-b±b2-4ac2ax=-b+b2-4ac2a or x=-b-b2-4ac2ax=-b2a+b2-4ac2a or x=-b2a-b2-4ac2aIf roots are imagianry then  b2-4ac<0.Let b2-4ac=iβ, where βx=-b2a+iβ2a or x=-b2a-iβ2aHence, we can see Rex=-b2aNote: b2a>0 as a,b>0-b2a<0Rex<0If roots are real then  b2-4ac0.x=-b2a+b2-4ac2a or x=-b2a-b2-4ac2ax=-b+b2-4ac2a or x=-b2a-b2-4ac2aNote: -b2a-b2-4ac2a<0 as a,b,c>0Also,Note:b2-4ac<b2b2-4ac<b2b2-4ac<b-b+b2-4ac<0-b+b2-4ac2a<0Hence, if x is real then both roots are<0Hence, Rex<0,both if roots are real and if roots are imaginary

  • 0
What are you looking for?