Applications of quadratic equations - class-X

Let $m$ be the age of Sam and $s$ be the age of his son
Then, 2 years ago,
$(m -2) = \dfrac{9}{2} (s-2)$
$2m - 4 = 9s - 18$
$2m = 9s - 14 ...(i)$
$6$ years ago,
$(m - 6) = 2(s- 6)^2$
$\left(\dfrac{9s-14}{2} - 6\right) = 2(s-6)^2$
$9s - 14 - 12 = 4(s^2 - 12s + 36)$
$9s - 26 = 4s^2 - 48s + 144$
$4s^2 - 57s + 170 = 0$
$s = \dfrac{57 \pm \sqrt{529}}{8}$ = $\dfrac{57 \pm 23}{8} = 10, 4.25$
Neglect the 4.25 which is a fraction. 
Hence, age of his son is $10$ years.

Let the time taken to complete the work  by B be  x.
A can take x - 6 hours.

A and B together can complete the work = 13 hours 20 minutes.
$\frac{1}{x-6}+\frac{1}{x}=13\frac{20}{60}$
$\frac{2x-6}{x^2-6x}=\frac{3}{40}$
$40(2x-6)=3x^2-18x$
$80x-240=3x^2-18x$
$3x^2-98x+240$
Using quadratic formula,
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\frac{98\pm\sqrt{(-98)^2-4\times 3\times 240}}{2\times 3}$
$\frac{98\pm\sqrt{6724}}{6}$
$\frac{98\pm 82}{6}$
So, x = 30, x = 2.6666..
x value cannot be negative.
B take 30 hours to complete the work alone.

Given A do a piece of work in x days and B do work in x+16 days

Let the present age of the son be $x$.
Let the father's present age be $y$.
One year ago, $y-1=8(x-1)$
$\Rightarrow  y-1=8x-8$
$\Rightarrow  y=8x-7$
Now applying the condition, we get
$(8x-7)=x^{2}$
$\Rightarrow x^{2}-8x+7=0$
$\Rightarrow (x-1)(x-7)=0$
$\Rightarrow x=1$ or $x=7$

Let the number of books he bought for Rs. $80$ be $x$

Let the number of students in each row be $x$.
Hence, $x(x+3)=460$
$x^{2}+3x=460$
$x^{2}+3x-460=0$
$(x+23)(x-20)=0$
$x=-23$ and $x=20$
Hence, the number students in  each column is $20+3=23$.

consider, $x-\cfrac{2}{x-1}=1-\cfrac{2}{x-1}$

Let y = 1/(x-2). Then x-2 = 1/y, so x = 2 + 1/y = (2y+1)/y. Substitute into x^2 - 3x + 1 = 0: ((2y+1)/y)^2 - 3((2y+1)/y) + 1 = 0. Simplifying leads to y^2 - y - 1 = 0.

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.

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

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

Let the original price of the books be $x$

Let the larger part be $x$

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

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 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

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

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

let present age of son be $x$

Given, $2\sqrt{x+5}=8$

${ 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 $

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.

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

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

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

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

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

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

$Clearly\quad x=1\quad is\quad a\quad solution$
$\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 } $
$\Rightarrow b-c=a-b$
$\Rightarrow 2b=a+c\Rightarrow a,b,c\quad are\quad in\quad A.P.$

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