Tag: applications of quadratic equations

Questions Related to applications of quadratic equations

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

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

  1. 0

  2. 1

  3. 2

  4. infinite

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Correct Option: A
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.

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

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

  1. $0$

  2. $\displaystyle \frac{1}{2}$

  3. $\displaystyle \frac{-1}{2}$

  4. $-1$

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Correct Option: D
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 

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

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

  1. $8$

  2. $7$

  3. $6$

  4. $5$

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Correct Option: A
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

$\therefore a=8$

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

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

  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

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Correct Option: C
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

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

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

$\therefore (\varphi+m)^2pr=\varphi mq^2$

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 

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

  1. $a >0$

  2. $a < 0$

  3. $\displaystyle (b^{2}-4ac)$ is negative

  4. Nothing can be said

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Correct Option: B
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 $a<0$

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.

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

  1. $15$

  2. $10$

  3. $20$

  4. $25$

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Correct Option: C
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, $x=20$

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

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

  1. $6, 10$

  2. $6, 4$

  3. $4, 10$

  4. None of these

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Correct Option: A
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 $10, 6$

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

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

  1. $10$ years

  2. $15$ years

  3. $20$ years

  4. $40$ years

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Correct Option: A
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

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.

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

  1. 18

  2. 72

  3. 27

  4. 81

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Correct Option: D
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 $\displaystyle \left( \frac { 80 }{ 1 } +1 \right) =81$.

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

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

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.

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Correct Option: C
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