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

The number of roots of the equation  $\displaystyle x-\frac{2}{(x-1)}=1-\frac{2}{(x-1)}$ is 

  1. 0

  2. 1

  3. 2

  4. infinite
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $ x - \cfrac {2}{(x-1)} = 1 - \cfrac {2}{(x-1)} $

Cancelling out $ - \cfrac {2}{(x-1)} $ from LHS and RHS we get, $ x = 1 $
But when $ x = 1 $, denominator of fraction $ - \cfrac {2}{(x-1)} $ is $ 0 $, which is not defined.
Hence, there is no root of this equation.

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

If $\displaystyle a^{2}+b^{2}+c^{2} = 1,$ then which of the following cannot be the value of $( ab + bc + ca)$?

  1. $0$
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{-1}{2}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$


$\Rightarrow (a+b+c) ^{2}=1+2(ab+bc+ca)$

$(a+b+c)^{2}$ should always be positive 

By checking options one by one, we get the result

1)ab+bc+ca=0

$\Rightarrow (a+b+c)^{2}=1$ which can be true 

Same way b, c options are true 

4th option where ab+bc+ca=-1/2

We get $(a+b+c)^{2}=-1$ which is false 

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

If $(x - a) (x - 5) + 2 = 0$ has only integral roots where $\displaystyle a \, \varepsilon \, I,$ then the value of $a$ can be 

  1. $8$
  2. $7$
  3. $6$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha, \beta $ be the roots of the given equation


$\alpha +\beta =5+a$

$\alpha ×\beta=5a+2$

Eliminating "a" from the equation, we get ;

$\Rightarrow 5(\alpha +\beta) - \alpha \beta =23$

Here we have to choose the pairs $\alpha, \beta$ so that it satisfies above equation 

By trial and error method, we find $\alpha =7,\beta=6$ or vice versa 

Substituting $\alpha, \beta$ in above equation, we get a=8

$\therefore a=8$

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

If the roots of the equation $\displaystyle px^{2}+qx+r=0$ are in the ratio $\displaystyle \varphi \ : \ m,$ then 

  1. $\displaystyle (\varphi +m)^{2}qp=\varphi mr^{2}$
  2. $\displaystyle (\varphi +m)^{2}pr=\varphi mq$
  3. $\displaystyle (\varphi +m)^{2}pr=\varphi mq^{2}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the roots be $\varphi r$ and $mr$
Sum of roots $= \varphi r+mr=-\dfrac{q}{p}$
$\therefore  (\varphi+m)r=-\dfrac{q}{p}$
$\therefore (\varphi+m)^2r^2=\dfrac{q^2}{p^2}$   ...(1)

Product of roots $= (\varphi r)(mr)=\dfrac{r}{p}$
$\therefore  \varphi mr^2=\dfrac{r}{p}$    ...(2)

Dividing equation (1) by (2), we get

$\dfrac{(\varphi+m)^2r^2}{\varphi mr^2}=\dfrac{\frac{q^2}{p^2}}{\frac{r}{p}}$

$\therefore \dfrac{(\varphi+m)^2}{\varphi m}=\dfrac{q^2}{pr}$

$\therefore (\varphi+m)^2pr=\varphi mq^2$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the graph of $|y| = f (x),$ where $\displaystyle f(x)=ax^{2}+bx+c; \ \ b \, & \, c \, \epsilon \, R; \ \ a\neq 0,$  has the maximum vertical height 4, then 

  1. $a >0$
  2. $a < 0$
  3. $\displaystyle (b^{2}-4ac)$ is negative

  4. Nothing can be said
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $:$ Graph of $|y|=f(x)$ , where $ax^2+bx+c$ has maximum vertical height 4

To find $:$ The condition of $a$
Solution $:$ All quadratic functions have a U-shaped graph called a parabola. 
The lowest or the highest point on a parabola is called the vertex. The vertex has the x-coordinate denoted by $x=-\dfrac b{2a}$
The y-coordinate of the vertex is the maximum or minimum value of the function.
If the y-coordinate has maximum value then parabola opens down, and $a<0$

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

If the list price of a book is reduced by Rs. $5$ a person can buy $5$ more books for Rs. $300$. Find the original list price of the book.

  1. $15$
  2. $10$
  3. $20$
  4. $25$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the original price of the books be $x$

$\therefore$ new price $=x-5$
Given, $\dfrac { 300 }{ x-5 } -\dfrac { 300 }{ x } =5$

$ 60x-60x+300={ x }^{ 2 }-5x$
$ { x }^{ 2 }-20x+15x-300=0$
$ (x-20)(x+15)=0$
$ x$ cannot be negative, i.e, $x=20$

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

Divide $16$ into two parts such that the twice of the square of the greater part exceeds the square of the smaller part by $164.$

  1. $6, 10$
  2. $6, 4$
  3. $4, 10$

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

Let the larger part be $x$

So, smaller part is $16-x$
Given, 
$2\times x^2 = \left(16-x\right)^2 + 164$
$2x^2 = \left(256-32x+x^2\right) +164$
$x^2 +32x -420 = 0$
$x^2 -10x +42x - 420 =0$
$ x \left(x-10 \right) +42 \left(x - 10 \right) = 0$
$\left(x - 10 \right) \left(x+42\right)$
$\rightarrow x = 10, -42$
$\because \ x$ cannot be negative, hence $x = 10$
So, the two parts are $10, 6$

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

Five years hence, father's age will be $3$ times the age of his son. Five years ago, father was seven times as old as his son. The age of the son at present is

  1. $10$ years
  2. $15$ years
  3. $20$ years
  4. $40$ years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $5$ years ago, the ages of son and father were $x$ and $7x$ years respectively, then 
$\displaystyle 3\left( x+5+5 \right) =7x+5+5$
$\displaystyle \Rightarrow  3x+30=7x+10$
$\displaystyle \Rightarrow  4x=20$
$\displaystyle \Rightarrow  x=5$
Thus, present age of son $= x + 5 = 10$ years

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

A two digit number in such that the product of its digits is $8$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.

  1. 18

  2. 72

  3. 27

  4. 81
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the digit at unit's place $= x$
$\displaystyle \therefore $ Digit at ten's place $\displaystyle \frac { 8 }{ x } $ and the number is $\displaystyle \left( \frac { 80 }{ x } +x \right) $
New number on interchanging the places of digits $\displaystyle =10x+\frac { 8 }{ x } $
$\displaystyle \therefore $ According to given condition
$\displaystyle \frac { 80 }{ x } +x-63=10x+\frac { 8 }{ x } $
$\displaystyle 80+{ x }^{ 2 }-63x=10{ x }^{ 2 }+8$
$\displaystyle { 9x }^{ 2 }+63x-72=0$
$\displaystyle { x }^{ 2 }-7x-8=0$
$\displaystyle { x }^{ 2 }+8x-x-8=0$
$\displaystyle x\left( x+8 \right) -1\left( x+8 \right) =0$
$\displaystyle \left( x+8 \right) \left( x-1 \right) =0$
$\displaystyle i.e.\quad x=-8$  and $ x=1$
Rejecting $\displaystyle x=-8$ and putting $\displaystyle x=1$ the required no. is $\displaystyle \left( \frac { 80 }{ 1 } +1 \right) =81$.

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

For the equation $|x|^{2}+|x|-6=0$, the roots are

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x>0 equation is $x^2+x-6=0$
$(x-2)(x+3)=0$
$x=2$
$x$ cant be equal to -3 as for this equation $x>0$
Now when $x <0$ equation becomes $x^2-x-6$
$(x+2)(x-3)=0$
Hence $x=-2$
So the roots are $2 and -2$
Thus sum of roots is zero and roots are real
So Option C