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

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Let $\alpha$ and $\beta$ be the roots of the equation $(x - a)(x - b) = c$,where $c \neq 0$. What are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

If $a > 0$,then the value of $\sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module