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

The number of values of $b$ for which there is an isosceles triangle with sides of lengths $b+5$,$3b-2$,and $6-b$ is

$A$ building construction work can be completed by two masons $A$ and $B$ together in $22.5$ days. Mason $A$ alone can complete the work in $24$ days less than mason $B$ alone. Then mason $A$ alone will complete the work in: (in $days$)

If $1$ is a multiple root of order $3$ for the equation $x^4-2x^3+2x-1=0$,then the other root is

The least positive value of $a$ for which the equation $2x^{2} + (a-10)x + \frac{33}{2} = 2a$ has real roots is

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module