Tag: order operations and algebra

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Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

If r and s are zeroes of the polynomial $t^2-4t+3$, then $\dfrac{1}{r}+\dfrac{1}{s}-2rs+\dfrac{14}{3}$ is equal to

  1. 0

  2. 1

  3. 2

  4. -1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Quadratic Equation
If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$
then $\alpha + \beta=\dfrac{-b}{a}$ and $\alpha\beta=\dfrac{c}{a}$

$f(t)=t^2-4t+3$
$r+s=4$
$rs=3$
Now,
$\dfrac{1}{r}+\dfrac{1}{s}-2rs+\dfrac{14}{3}=\dfrac{r+s}{rs}-2rs+\dfrac{14}{3}$

$=\dfrac{4}{3}-6+\dfrac{14}{3}=\dfrac{4-18+14}{3}=0$
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

The simplified form of the expression given below is :$\dfrac{\dfrac{y^4-x^4}{x(x+y)}-\dfrac{y^3}{x}}{y^2-xy+x^2}$

  1. $1$
  2. $0$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given expression $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$ can be simplified as follows:

 
$\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 })^{ 2 }-({ x }^{ 2 })^{ 2 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 }-{ x }^{ 2 })({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \quad \quad \quad \quad \quad \quad \quad \left( \because \quad a^{ 2 }-b^{ 2 }=(a+b)(a-b) \right)$
$=\dfrac { \dfrac { ({ y }+x)(y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { (y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { { y }^{ 3 }-xy^{ 2 }+yx^{ 2 }-{ x }^{ 3 }-{ y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$
$=\dfrac { \dfrac { -x(y^{ 2 }-xy+{ x }^{ 2 }) }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-\dfrac { y^{ 2 }-xy+{ x }^{ 2 } }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-1$

Hence, $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }=-1$  

Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

In the equation $4x+y=10$, if the value of $x$ ins increased by $3$, then what would be the effect on the corresponding value of $y$

  1. The value of $y$ is decreased by $12$
  2. The value of $y$ is decreased by $2$
  3. The value of $y$ is increased by $3$
  4. The value of $y$ will be $3$ times as large
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 4x + y = 10, if x increases by 3, the new equation is 4(x + 3) + y_new = 10. This expands to 4x + 12 + y_new = 10, which simplifies to y_new = 10 - 4x - 12. Since the original y = 10 - 4x, the new y is y - 12.

Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

Evaluate:
$( b - c + d + a ) ( d + a - b + c ) + ( c - d + a + b ) ( b + c + d - a )$

  1. $4 ( a d + b c )$
  2. $2 ( a d + b c )$
  3. $3 ( a d + b c )$
  4. $ ( a d + b c )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(b-c+d+a)(d+a-b+c)+(c-d+a+b)(b+c+d-a)$

$=[(d+a)+(b-c)][(d+a)-(b-c)]+[(b+c)+(a-d)][(b+c)-(a-d)]$

$[\because (a-d)^2=(d-a)^2]$

$=(d+a)^2-(b-c)^2+(b+c)^2-(a-d)^2$

$=(d+a)^2-(d-a)^2+(b+c)^2-(b-c)^2$

$=4ad+4bc$

$=4(ad+bc)$.