The roots of the equation ({a^2} + {b^2}){t^2 | vedclass

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

Question

(D) For a quadratic equation $At^2 + Bt + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = (a^2 + b^2)$,$B =

Vedclass · Accepted Answer

$\frac{a}{b} = \frac{c}{d}$

Vedclass · Answer

(D) For a quadratic equation $At^2 + Bt + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = (a^2 + b^2)$,$B = -2(ac + bd)$,and $C = (c^2 + d^2)$.Setting $D = 0$ gives:$[-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$$4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$$(ac + bd)^2 - (a^2 + b^2)(c^2 + d^2) = 0$$a^2c^2 + b^2d^2 + 2abcd - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$$a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2 = 0$$-a^2d^2 - b^2c^2 + 2abcd = 0$$-(a^2d^2 + b^2c^2 - 2abcd) = 0$$-(ad - bc)^2 = 0$$(ad - bc)^2 = 0$$ad = bc$$\frac{a}{b} = \frac{c}{d}$.

The roots of the equation $({a^2} + {b^2}){t^2} - 2(ac + bd)t + ({c^2} + {d^2}) = 0$ are equal. Then:

Similar Questions

Let $a_n, a_{n-1}, \ldots, a_1, a_0 \in \mathbb{C}$ and $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ be a polynomial. If the polynomial $f(x)$ is monic,then:

If the roots of the equations $ax^2 + 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real,then:

Solve the equation $21x^{2} - 28x + 10 = 0$.

Consider the equation $x^2+x-n = 0$,where $n \in N$ and $n \in [5, 100]$. The total number of different values of $n$ such that the given equation has integral roots is:

The sum of the non-real roots of $(p^2+p-3)(p^2+p-2)-12=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module