If {b_1}{b_2} = 2({c_1} + {c_2}),then at leas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let ${D_1}$ and ${D_2}$ be the discriminants of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ respectively.Then,${D_

Vedclass · Accepted Answer

Real roots

Vedclass · Answer

(A) Let ${D_1}$ and ${D_2}$ be the discriminants of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ respectively.Then,${D_1} = b_1^2 - 4{c_1}$ and ${D_2} = b_2^2 - 4{c_2}$.Adding the two discriminants,we get:${D_1} + {D_2} = b_1^2 + b_2^2 - 4({c_1} + {c_2})$.Given that ${b_1}{b_2} = 2({c_1} + {c_2})$,we can substitute $4({c_1} + {c_2}) = 2{b_1}{b_2}$ into the equation:${D_1} + {D_2} = b_1^2 + b_2^2 - 2{b_1}{b_2} = {(b_1 - b_2)^2}$.Since the square of any real number is non-negative,${(b_1 - b_2)^2} \ge 0$,which implies ${D_1} + {D_2} \ge 0$.If the sum of two numbers is non-negative,at least one of them must be non-negative.Therefore,${D_1} \ge 0$ or ${D_2} \ge 0$,which means at least one of the equations has real roots.

If ${b_1}{b_2} = 2({c_1} + {c_2})$,then at least one of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ has

Similar Questions

Which of the following conditions implies that the roots of the equation $\frac{1}{4}x^2 + bx + c = 0$ are integers?

The number of distinct real solutions for the equation $|x^2+2x-8|+x-2=0$ is

If the equation $x^2 - m(2x - 8) - 15 = 0$ has equal roots,then $m = ......$

If $\sinh(\log x) = -2$,then $x =$

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4-4x^3+3x^2+2x-2=0$ such that $\alpha$ and $\beta$ are integers and $\gamma, \delta$ are irrational numbers,then $\alpha+2\beta+\gamma^2+\delta^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module