If the roots of the equation ({p^2} + {q^2}){ | vedclass

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

Question

(B) Given the quadratic equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.Since the roots are real and equal,the discriminant $D = b^

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(B) Given the quadratic equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.Since the roots are real and equal,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = ({p^2} + {q^2})$,$b = -2q(p + r)$,and $c = ({q^2} + {r^2})$.Substituting these into $D = 0$:$[-2q(p + r)]^2 - 4({p^2} + {q^2})({q^2} + {r^2}) = 0$$4q^2(p + r)^2 - 4({p^2}q^2 + p^2r^2 + q^4 + q^2r^2) = 0$Dividing by $4$:$q^2(p^2 + r^2 + 2pr) - (p^2q^2 + p^2r^2 + q^4 + q^2r^2) = 0$$p^2q^2 + q^2r^2 + 2pq^2r - p^2q^2 - p^2r^2 - q^4 - q^2r^2 = 0$Simplifying the expression:$2pq^2r - p^2r^2 - q^4 = 0$$-(q^4 - 2pq^2r + p^2r^2) = 0$$-(q^2 - pr)^2 = 0$$(q^2 - pr)^2 = 0$This implies $q^2 = pr$.Therefore,$p, q, r$ are in $G.P.$ (Geometric Progression).

If the roots of the equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$ are real and equal,then $p, q, r$ will be in

Similar Questions

If the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ will be

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c,$ where $c \neq 0,$ then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

$\alpha_r$ and $\beta_r$ $(\alpha_r < \beta_r)$ are the roots of $x^2 - r^2(r + 1)x + r^5 = 0$. The value of $\sum_{r=1}^n (3\alpha_r + 2\beta_r)$ is:

Solve the given two equations and choose the correct answer from the given options.
$I.$ $16x^2 + 20x + 6 = 0$
$II.$ $10y^2 + 38y + 24 = 0$

Solve the given two equations and select the correct answer from the given options:
$I.$ $\sqrt{1225} x + \sqrt{4900} = 0$
$II.$ $(81)^{1/4} y + (343)^{1/3} = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module