Finding roots or coefficients of a quadratic using Vieta's relations
The question asks to find roots, coefficients, or a related quadratic equation by applying relationships between roots and coefficients (sum and product of roots) rather than directly solving the equation.
Question 173 Um número inteiro positivo de dois algarismos é tal que a soma de seus algarismos é 9 e o produto de seus algarismos é 18. Esse número é (A) 27 (B) 36 (C) 45 (D) 63 (E) 72
8. If $a$ and $b$ are two distinct zeros of the function $f ( x ) = x ^ { 2 } - p x + q$ ($p > 0$, $q > 0$), and the three numbers $a$, $b$, and $-2$ can be arranged to form an arithmetic sequence, and can also be arranged to form a geometric sequence, then the value of $p + q$ equals A. 6 B. 7 C. 8 D. 9
Suppose $a , b$ denote the distinct real roots of the quadratic polynomial $x ^ { 2 } + 20 x - 2020$ and suppose $c , d$ denote the distinct complex roots of the quadratic polynomial $x ^ { 2 } - 20 x + 2020$. Then the value of $$a c ( a - c ) + a d ( a - d ) + b c ( b - c ) + b d ( b - d )$$ is (A) 0 (B) 8000 (C) 8080 (D) 16000
If equations $a x ^ { 2 } + b x + c = 0 , ( a , b , c \in R , a \neq 0 )$ and $2 x ^ { 2 } + 3 x + 4 = 0$ have a common root, then $a : b : c$ equals : (1) $2 : 3 : 4$ (2) $4 : 3 : 2$ (3) $1 : 2 : 3$ (4) $3 : 2 : 1$
Let $p , q$ and $r$ be real numbers ( $p \neq q , r \neq 0$ ), such that the roots of the equation $\frac { 1 } { x + p } + \frac { 1 } { x + q } = \frac { 1 } { r }$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to (1) $p ^ { 2 } + q ^ { 2 }$ (2) $\frac { p ^ { 2 } + q ^ { 2 } } { 2 }$ (3) $2 \left( p ^ { 2 } + q ^ { 2 } \right)$ (4) $p ^ { 2 } + q ^ { 2 } + r ^ { 2 }$
Let $a , b \in R , a \neq 0$ be such that the equation, $a x ^ { 2 } - 2 b x + 5 = 0$ has a repeated root $\alpha$, which is also a root of the equation, $x ^ { 2 } - 2 b x - 10 = 0$. If $\beta$ is the other root of this equation, then $\alpha ^ { 2 } + \beta ^ { 2 }$ is equal to: (1) 25 (2) 26 (3) 28 (4) 24
Let $\lambda \neq 0$ be in $R$. If $\alpha$ and $\beta$ are the roots of the equation, $x ^ { 2 } - x + 2 \lambda = 0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x ^ { 2 } - 10 x + 27 \lambda = 0$, then $\frac { \beta \gamma } { \lambda }$ is equal to: (1) 27 (2) 18 (3) 9 (4) 36
If $\alpha$ and $\beta$ are the roots of the equation $2\mathrm{x}(2\mathrm{x}+1)=1$, then $\beta$ is equal to: (1) $2\alpha(\alpha+1)$ (2) $-2\alpha(\alpha+1)$ (3) $2\alpha(\alpha-1)$ (4) $2\alpha^{2}$
Let $a, b \in R$ be such that the equation $ax^2 - 2bx + 15 = 0$ has repeated root $\alpha$ and if $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2bx + 21 = 0$, then $\alpha^2 + \beta^2$ is equal to: (1) 37 (2) 58 (3) 68 (4) 92
If for some $p , q , r \in R$, all have positive sign, one of the roots of the equation $\left( p ^ { 2 } + q ^ { 2 } \right) x ^ { 2 } - 2 q ( p + r ) x + q ^ { 2 } + r ^ { 2 } = 0$ is also a root of the equation $x ^ { 2 } + 2 x - 8 = 0$, then $\frac { q ^ { 2 } + r ^ { 2 } } { p ^ { 2 } }$ is equal to $\_\_\_\_$.
If the value of real number $\alpha > 0$ for which $x^{2} - 5\alpha x + 1 = 0$ and $x^{2} - \alpha x - 5 = 0$ have a common real roots is $\frac{3}{\sqrt{2\beta}}$ then $\beta$ is equal to $\_\_\_\_$
Let $x = \frac { m } { n } \left( m , n \right.$ are co-prime natural numbers) be a solution of the equation $\cos \left( 2 \sin ^ { - 1 } x \right) = \frac { 1 } { 9 }$ and let $\alpha , \beta ( \alpha > \beta )$ be the roots of the equation $m x ^ { 2 } - n x - m + n = 0$. Then the point $( \alpha , \beta )$ lies on the line (1) $3 x + 2 y = 2$ (2) $5 x - 8 y = - 9$ (3) $3 x - 2 y = - 2$ (4) $5 x + 8 y = 9$
Let $\alpha _ { \theta }$ and $\beta _ { \theta }$ be the distinct roots of $2 x ^ { 2 } + ( \cos \theta ) x - 1 = 0 , \theta \in ( 0,2 \pi )$. If m and M are the minimum and the maximum values of $\alpha _ { \theta } ^ { 4 } + \beta _ { \theta } ^ { 4 }$, then $16 ( M + m )$ equals : (1) 24 (2) 25 (3) 17 (4) 27
Let $a, b, c$ be nonzero real numbers, and the two roots of the quadratic equation $ax^2 + bx + c = 0$ both lie between 1 and 3. Select the equation whose two roots must lie between 4 and 5. (1) $a(x-2)^2 + b(x-2) + c = 0$ (2) $a(x+2)^2 + b(x+2) + c = 0$ (3) $a(2x-7)^2 + b(2x-7) + c = 0$ (4) $a\left(\frac{x+7}{2}\right)^2 + b\left(\frac{x+7}{2}\right) + c = 0$ (5) $a(3x-11)^2 + b(3x-11) + c = 0$
$$(3x-1)(x+1)+(3x-1)(x-2)=0$$ What is the sum of the real numbers $x$ that satisfy the equation? A) $\frac{2}{3}$ B) $\frac{3}{4}$ C) $\frac{3}{5}$ D) $\frac{5}{6}$ E) $\frac{7}{6}$