Tag: applications of quadratic equations

Quizzes (0) Questions (39)

Questions Related to applications of quadratic equations

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x^{2} + 2xy + 3xz = 50$,
$2y^{2} + 3yz + yx = 10$,
$3z^{2} + zx + 2zy = 10$.

  1. $x=\pm 4; y=\pm 2; z=\pm 2$
  2. $x=\pm -4; y=\pm -2; z=\pm 2$
  3. $x = \pm 5; y = \pm 1; z = \pm 1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given equations are ${ x }^{ 2 }+2xy+3xz=50$

$\Rightarrow  x(x+2y+3z)=50$    ........(i),
$ 2{ y }^{ 2 }+3yz+yx=10$
$\Rightarrow  y(2y+3z+x)=10$    ........(ii)
and $ 3{ z }^{ 2 }+zx+2zy=10$
$\Rightarrow  z(3z+x+2y)=10$    ..........(iii)
Dividing (i) by (ii), we get

$\dfrac { x }{ y } =\dfrac { 50 }{ 10 } =5\\ \Rightarrow x=5y$    ......(a)
Dividing (ii) by (iii), we get
$\dfrac { y }{ z } =\dfrac { 10 }{ 10 } =1\\ \Rightarrow z=y$    ..........(b)
Substituting (a) and (b) in (ii)
$y(2y+3y+5y)=10\\ \Rightarrow 10{ y }^{ 2 }=10\\ \Rightarrow y=\pm 1$
From (a), we have
$x=\pm 5$
From (b), we have
$z=\pm 1$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x + 2y - z = 11$,
$x^{2} - 4y^{2} + z^{2} = 37$,
$xz = 24$.

  1. $x=2, -5; y=2; z=2,-4$
  2. $x=8,-3; y=3; z=-3,-8$
  3. $x=-3, 5; y=4; z=2,5$
  4. $x=2,4; y=3; z=3, -5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ x }^{ 2 }-4{ y }^{ 2 }+{ z }^{ 2 }=37$    ......(i)

$xz=24$    .......(ii)
$ x+2y-z=11$    .....(iii)
$\Rightarrow  x-z=11-2y$
On squaring both sides, we have

${ x }^{ 2 }+{ z }^{ 2 }-2xz=121+4{ y }^{ 2 }-44y\\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }-4{ y }^{ 2 }-2xz=121-44y\\ \Rightarrow 37-2(24)=121-44y\\ \Rightarrow -44y=-132\\ \Rightarrow y=3$
Substituting $y$ in (iii), 
$x+6-z=11\\ \Rightarrow x-z=5$
From (ii), $z=\dfrac { 24 }{ x } $
Thus $x-\dfrac { 24 }{ x } =5$
$ \Rightarrow { x }^{ 2 }-24=5x\\ \Rightarrow { x }^{ 2 }-5x-24=0\\ \Rightarrow { x }^{ 2 }-3x+8x-24=0\\ \Rightarrow x(x-3)+8(x-3)=0\\ \Rightarrow (x+8)(x-3)=0\\ \Rightarrow x=-8,3$
Putting in $ z=\dfrac { 24 }{ x } $
Thus $ z=-3,8$
So, the values of $x$ are $-8,3$, values of $z$ are $-3,8$ and value of $y$ is $3$.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the zeroes of the rational expression $ (ax+b)(3x+2)$ are $-\dfrac{2}{3}$ and $ \dfrac{1}{2}$, then $ a+b=$

  1. $4$
  2. $0$
  3. $-b$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(ax+b)(3x+2)=0$

$\Rightarrow$  $3ax^2+2ax+3bx+2b$
$\Rightarrow$  $3ax^2+(2a+3b)x+2b=0$
It is given that $\dfrac{-2}{3}$ and $\dfrac{1}{2}$ are zeros of the equation $3ax^2+(2a+3b)x+2b=0$
Here, $A=3a,\,B=2a+3b,\,C=2b$
Let $\alpha=\dfrac{-2}{3}$ and $\beta=\dfrac{1}{2}$
We know,
$\Rightarrow$  $\alpha+\beta=\dfrac{-B}{A}$
$\Rightarrow$  $\dfrac{-2}{3}+\dfrac{1}{2}=\dfrac{-(2a+3b)}{3a}$

$\Rightarrow$  $\dfrac{-4+3}{6}\times 3a=-2a-3b$

$\Rightarrow$  $\dfrac{-a}{2}=-2a-3b$
$\Rightarrow$  $-a=-4a-6b$
$\Rightarrow$  $3a+6b=0$              ----- ( 1 )
Now,
$\Rightarrow$  $\alpha.\beta=\dfrac{CA}{A}$

$\Rightarrow$  $\dfrac{-2}{3}\times\dfrac{1}{2}=\dfrac{2b}{3a}$

$\Rightarrow$  $\dfrac{-1}{3}\times 3a=2b$

$\Rightarrow$  $-a+2b=0$          ----- ( 2 )
Adding equation ( 1 ) and ( 2 ) we get,
$b=0$
Put $b=0$ in ( 2 ) we get,
$a=0$
$\therefore$  $a+b=0+0=0$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

In a bangle shop, if the shopkeeper displays the bangles in the form of a square then he is left with 38 bangles with him. If he wanted to increase the size of square by one unit each side of the square he found that 25 bangles fall short of In completing the square. The actual number of bangles which he had with him in the shop was ________.

  1. 1690

  2. 999

  3. 538

  4. can't be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the length of the side of the square initially be $l$.

Number of bangles it can store is $l^{2}$
The total number of bangles will be $l^{2}+38$
In the second case,
The length of the side of the square is $l+1$
The number of bangles it can now store is $(l+1)^{2}$
Now the total number of bangles equals to $(l+1)^{2}-25$
$\therefore$ $(l+1)^{2}-25$=$l^{2}+38$
$\Rightarrow$ $l^{2}+2l+1-25=l^{2}+38$
$\Rightarrow$ $2l=62$
$\Rightarrow$ $l=31$
$\therefore$ The total number of bangles=$(l+1)^{2}-25$=$999$
$\therefore$ Option B is correct

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

A man walks a distance of 48 km in a given time. If he walks 2 km/hr faster, he will perform the journey 4 his before. His normal rate of walking is _______.

  1. 3 km/hr

  2. 4 km/hr
    • 6 km/hr or 4 km/hr

  4. 5 km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let us assume that the man walks at a speed of $x$ $kmph$

Then the time taken by the man to complete the journey is $t=\frac{48}{x}$  ($\because time=\frac {distance}{speed}$)
Now in the second case, $t'=t-3$ and $x'=x+2$.
Then, 
$t-4=\frac{48}{x+2}$
$\Rightarrow$ $t-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{48}{x}-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{96}{x(x+2)}=4$
$\Rightarrow$ $x^{2}+2x=24$
$\Rightarrow$ $x^{2}+2x-24=0$
$\Rightarrow$ $x^{2}-4x+6x-24=0$
$\Rightarrow$ $(x-4)(x+6)=0$
Speed cannot be a negative value and hence $x=4kmph$
$\therefore$ Option B is correct.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

 Choose the correct answer from the alternatives given.
If $\alpha \, and \, \beta$ are the roots of the equation $x^2$ - 7x + 12 = 0, then $\alpha^2 \, + \, \beta^2$ equals.

  1. 19

  2. 25

  3. 14

  4. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
: Let $\alpha \, and \, \beta$ are the roots of the equation
$ax^2 + bx + c = 0$
We know that,
$\displaystyle \alpha \, + \, \beta \, = \, \frac{-b}{a} \, = \, \frac{- (-7)}{1} \, = \, 7$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2 \, - \, 2\alpha \beta$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (7)^2 \, - \, 2 \, \times \, 12 \, = \, 49 \, - \, 24 \, = \, 25$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Their present ages are 6 years and 12 years.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  Let present age of girl be $2x$.

$\Rightarrow$  Then, age of her sister be $x.$
According to the question,
$\Rightarrow$  $(2x+4)(x+4)=160$
$\Rightarrow$  $2x^2+8x+4x+16=160$
$\Rightarrow$  $2x^2+12x+16-160=0$
$\Rightarrow$  $2x^2+12x-144=0$
$\Rightarrow$  $2(x^2+6x-42)=0$
$\Rightarrow$  $x^2+6x-72=0$
$\Rightarrow$  $x^2+12x-6x-72=0$
$\Rightarrow$  $x(x+12)-6(x+12)=0$
$\Rightarrow$  $(x+12)(x-6)=0$
$\Rightarrow$  $x+12=0$ and $x-6=0$
$\therefore$  $x=-12$ and $x=6$
$\Rightarrow$  Age cannot be negative.
$\therefore$  $x=6$
$\Rightarrow$  $2x=2\times 6=12$
$\therefore$  The present ages of girls are $6\,years$ and $12\,years.$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If  a,b,c are distinct and the roots of $\left( b-c \right) { x }^{ 2 }+\left( c-a \right) x+(a-b)=0$ are equal, then a,b,c are in

  1. Arithmetic progression

  2. Geometric progression

  3. Harmonic progression

  4. Arithmetico-Geometric progression
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Clearly\quad x=1\quad is\quad a\quad solution$
$\therefore \quad Product\quad of\quad the\quad roots=\frac { a-b }{ b-c } $$\therefore \quad \left( 1 \right) \left( 1 \right) =\frac { a-b }{ b-c } $
$\Rightarrow b-c=a-b$
$\Rightarrow 2b=a+c\Rightarrow a,b,c\quad are\quad in\quad A.P.$






Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the harmonic mean of the roots of$\sqrt { 2 } { x }^{ 2 }-bx+\left( 8-2\sqrt { 5 }  \right) =0$ is 4, the the value of b=

  1. 2

  2. 3
  3. $4-\sqrt { 5 } $
  4. $4+\sqrt { 5 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Let\quad \alpha ,\beta \quad be\quad the\quad roots$ 
$\Rightarrow \frac { 2\alpha \beta  }{ \alpha +\beta  } =4\quad \quad \Rightarrow \frac { \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ \sqrt { 2 }  }  }{ \frac { b }{ \sqrt { 2 }  }  } =4$
$\Rightarrow \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ 4 } =b$
$\therefore b=4-\sqrt { 5 } $