Tag: word problems based on quadratic equations

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

If $a, b, c$ are in A.P., then the roots of the equation $ax^{2}+2bx+c=0$ are

  1. real and distinct

  2. real and equal

  3. real

  4. imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $a,b,c$ are in A.P, then

$2b=a+c$.
Hence the give equation transforms into
$ax^{2}+(a+c)x+c=0$
Hence
$D$
$=B^{2}-4AC$

$=(a+c)^{2}-4ac$

$=a^{2}+c^{2}+2ac-4ac$

$=a^{2}+c^{2}-2ac$

$=(a-c)^{2}$
Now 
$(a-c)^{2}\geq 0$
Hence 
$D\geq 0$.
Or 
$B^{2}-4AC\geq 0$.
Since the discriminant is greater than 0, hence the roots real.

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

To fill a cistern, pipes $P, Q$ & $R$ take $20, 15$ and $12$ minutes respectively.  The time in minutes that the three pipes together will take to fill the cistern is 

  1. $5$ min
  2. $10$ min
  3. $15$ min
  4. $15.66$ min
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: pipes $P, Q, R$ takes $20, 15, 12$ minutes respectively to fill a cistern

To find the time in minutes that the three pipes together will take to fill the cistern
Sol: By taking the LCM of $12, 15,20$ we will be able to find the actual capacity of the cistern, i.e., the actual capacity of cistern is 60 L(lets assume the unit to be litres)
IF pipe P can fill the cistern in $20$ minutes. Then in 1 min it can fill $\dfrac {60}{20}=3$ liters

Similarly in 1 min pipe Q can fill $\dfrac {60}{15}=4$ liters
And pipe R can fill $\dfrac {60}{12}=5$ liters
Therefore in 1 min the cistern will be filled by (P in 1 min)+(Q in 1 min)+(R in 1 min)= $(3+4+5) =12$ liters
Now, 
Time taken to fill $12 $ liters=$1$ minute
Therefore, 
Time taken to fill $60$ litres=$\dfrac {60}{12}$ (apply unitary method) =$5$ mins
Therefore, all the three pipes can fill the cistern together in 5 mins

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

A ball is thrown upwards from a rooftop, $28$m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time $t$ is $h$, which is given by, $h =$ $(-t^{2}+2t +35) +28$

How long will it take before hitting the ground?

  1. $6$ seconds
  2. $7$ seconds
  3. $8$ seconds
  4. $9$ seconds
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$h =$ $-t^{2}+2t +35 + 28=0$
$=>$ $t^{2}-2t - 63=0$
$=>(t - 9)(t + 7) = 0$
$=>t = 9$ or $-7$
The time cannot be negative, so the time is $9$ seconds.

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

The age of a man is the square of his son's age. A year ago, the man's age was $8$ times the age of his son. What is the present age of the man?

  1. $47\ yr$
  2. $49\ yr$
  3. $36\ yr$
  4. $48\ yr$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

let present age of son be $x$

present age of father be ${ x }^{ 2 }$
$1$ year ago
age of son $=x-1$
age of father = ${ x }^{ 2 }$-1
so 
${ x }^{ 2 }-1=8(x-1)$
${ x }^{ 2 }-8x+7=0$
$(x-7)(x-1)=0$
$x=7$
son's age $= 7 years$
man's age = ${ x }^{ 2 }=49$ years

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

If $a$ and $b$ are the roots of the quadratic equation $x^2-4x+3=0$, then $(1+a+a^2+a^3...)(1+b+b^2+b^3+....)$ equal to

  1. $\infty$
  2. $\dfrac{1}{4}$
  3. $\dfrac{-1}{6}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ x }^{ 2 }-4x+3=0\ (x-1)(x-3)=0\ x=1\quad or\quad 3\ \therefore a=3,\quad b=1\ a+b=4,\quad ab=3\ 1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty =\cfrac { 1 }{ 1-a } (sum\quad of\quad infinite\quad G.P.)\ \therefore (1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty )(1+b+{ b }^{ 2 }+{ b }^{ 3 }+.......+\infty )\ =(\cfrac { 1 }{ 1-a } )(\cfrac { 1 }{ 1-b } )\ =\cfrac { 1 }{ 1-(a+b)+ab } \ =\cfrac { 1 }{ 1-4+3 } =\cfrac { 1 }{ 0 } =\infty $

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

The number of real solution of the equation $(\dfrac{9}{10})^x=-3+x-x^2$ is-

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $f\left( x \right) ={ \left( \cfrac { 9 }{ 10 }  \right)  }^{ x }$
$g\left( x \right) =-3+x-{ x }^{ 2 }$
For $x<0$
$f\left( x \right) >1$
$g\left( x \right) -1=-{ x }^{ 2 }+x-4$
$\triangle =1-4\left( -1 \right) \left( -4 \right) $
$=-15<0$
and co efficient of ${ x }^{ 2 }=-1<0$
$\therefore g\left( x \right) -1<0$
$\therefore g\left( x \right) <1$
$\therefore f\left( x \right) -g\left( x \right) $ has no solution if $x<0$
For $x>0$
$f\left( x \right) <1$
$-3+x-{ x }^{ 2 }=f\left( x \right) $
$-{ x }^{ 2 }+x-3-f\left( x \right) =0$
$\triangle =1-4\left( -1 \right) \left( -3-f\left( x \right)  \right) $
$=1-12+4f\left( x \right) $
$\triangle =-11+4f\left( x \right) $
$f\left( x \right)<0$
$4f\left( x \right)-11<-11$
$\therefore \triangle <0$
$f\left( x \right)=g\left( x \right)$ has no real solution for $x>0$
At $x=0$
$f\left( x \right)=1,g\left( x \right)=-3$
$\therefore f\left( x \right)=g\left( x \right)$ has no real solution $\forall x\epsilon R$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $a,b,c,d$ are four consecutive terms of an increasing A.P., then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are

  1. $\text{real and distinct}$
  2. $\text {non-real complex}$
  3. $\text {real and equal}$
  4. $\text {integers}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a,b,c,d$ are $4$ consecutive terms of A

Let $a=m-3n,b=m-n.c=m+n,d=m+3n$

$'2n'$ is common difference

$(x-a)(x-c)+2(x-b)(x-d)=0$

$3{ x }^{ 2 }-\left( a+c+2b+2d \right) x+\left( ac+2bd \right) =0$

$=6m+2n$

$ac+2bd={ m }^{ 2 }-2mn-3{ n }^{ 2 }+2{ m }^{ 2 }+4mn-6{ n }^{ 2 }$

$=3{ m }^{ 2 }+2mn-9{ n }^{ 2 }$

$3{ x }^{ 2 }-\left( 6m+2n \right) x+\left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) =0$

$\triangle ={ \left( 6m+2n \right)  }^{ 2 }-4\left( 3 \right) \left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) $

$=4\left( 9{ m }^{ 2 }+6mn+{ n }^{ 2 }-9{ m }^{ 2 }-6mn+36{ n }^{ 2 } \right) $

$=4\left( 37 \right) { n }^{ 2 }$

$\triangle >0$

$\therefore $ Roots of $(x-a)(x-c)+2(x-b)(x-d)$ are real and distinct.
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If roots of equation $ x^2 - (2n+ 18) x - n-1 = 0 ( n \epsilon Z ) $ are rational, then number of possible value of $n $ is :

  1. $1$
  2. $2$
  3. $0$

  4. Infinite
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be rational, the discriminant D = (2n+18)^2 - 4(1)(-n-1) must be a perfect square. D = 4n^2 + 72n + 324 + 4n + 4 = 4n^2 + 76n + 328. Setting 4n^2 + 76n + 328 = k^2, we find integer solutions for n.