Algebra Questions

Multiple choice
  1. $x_1=1,x_2=2$
  2. $x+1=2,x+2=3$
  3. $x_1=3,x_2=4$
  4. $x_1=5,x_2=6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The quadratic equation x^2 - 7x + 12 = 0 can be factored as (x-3)(x-4) = 0. Thus, the roots are x = 3 and x = 4.

Multiple choice
  1. If $x _ { 1 }$ $\&$ $x _ { 2 }$ are roots of the equation $2 x ^ { 2 } - 6 x - b = 0 ( b > 0 ) ,$ then
    $\dfrac { x _ { 1 } } { x _ { 2 } } + \dfrac { x _ { 2 } } { x _ { 1 } } < - 2$
  2. Equation $a x ^ { 2 } + b x + c = 0$ has real roots if $a < 0 , c > 0$ and $b \in R$
  3. If $P ( x ) = a x ^ { 2 } + b x + c$ and $Q ( x ) = - a x ^ { 2 } + b x + c ,$ where $a c \neq 0$ and ${ a },{ b },{ c }\in { R },$ then ${ P }({ x })\cdot { Q }({ x })$ has at least two real roots.
  4. None of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice
  1. $log\left( { x }^{ 2 }+px+q \right) $
  2. $log\left( { x }^{ 2 }-px+q \right) $
  3. $log\left( 1+px+{ qx }^{ 2 } \right) $
  4. $log\left( 1-px+{ qx }^{ 2 } \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice
  1. $\frac { \sqrt { 5 } +1 }{ 2 } $
  2. $\frac { \sqrt { 5 } -1 }{ 2 } $
  3. 1

  4. 2

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let alpha be the common root. Then a*alpha^2 + b*alpha + c = 0 and c*alpha^2 + b*alpha + a = 0. Subtracting gives (a-c)(alpha^2 - 1) = 0. If a != c, alpha = 1 or -1. If alpha = 1, a+b+c=0. If alpha = -1, a-b+c=0. Using the difference of other roots = 1, the maximum value of |a/c| is (sqrt(5)+1)/2.