Solution

(i)

$ \begin{aligned} & x^{2}-3 x-10 \\ & =x^{2}-5 x+2 x-10 \\ & =x(x-5)+2(x-5) \\ & =(x-5)(x+2) \end{aligned} $

Roots of this equation are the values for which $(x-5)(x+2)=0$ $\therefore x-5=0$ or $x+2=0$

i.e., $x=5$ or $x=-2$

(ii) $2 x^{2}+x-6$

$ \begin{aligned} & =2 x^{2}+4 x-3 x-6 \\ & =2 x(x+2)-3(x+2) \\ & =(x+2)(2 x-3) \end{aligned} $

Roots of this equation are the values for which $(x+2)(2 x-3)=0$

$\therefore x+2=0$ or $2 x-3=0$

i.e., $x=-2$ or $x=\dfrac{3}{2}$

(iii) $\sqrt{2} x^{2}+7 x+5 \sqrt{2}$

$=\sqrt{2} x^{2}+5 x+2 x+5 \sqrt{2}$

$=x(\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)$

$=(\sqrt{2} x+5)(x+\sqrt{2})$

Roots of this equation are the values for which $(\sqrt{2} x+5)(x+\sqrt{2})=0$

$\therefore \sqrt{2} x+5=0$ or $x+\sqrt{2}=0$

i.e., $x=\dfrac{-5}{\sqrt{2}}$ or $x=-\sqrt{2}$

(iv) $2 x^{2}-x+\dfrac{1}{8}$

$=\dfrac{1}{8}(16 x^{2}-8 x+1)$

$=\dfrac{1}{8}(16 x^{2}-4 x-4 x+1)$

$=\dfrac{1}{8}(4 x(4 x-1)-1(4 x-1))$

$=\dfrac{1}{8}(4 x-1)^{2}$

Roots of this equation are the values for which $(4 x-1)^{2}=0$

Therefore, $(4 x-1)=0$ or $(4 x-1)=0$

i.e., $x=\dfrac{1}{4}$ or $x=\dfrac{1}{4}$

(v) $100 x^{2}-20 x+1$

$=100 x^{2}-10 x-10 x+1$

$=10 x(10 x-1)-1(10 x-1)$

$=(10 x-1)^{2}$

Roots of this equation are the values for which $(10 x-1)^{2}=0$

Therefore, $(10 x-1)=0$ or $(10 x-1)=0$

i.e., $x=\dfrac{1}{10}$ or $x=\dfrac{1}{10}$

Solution

(i) Let the number of John’s marbles be $x$.

Therefore, number of Jivanti’s marble $=45-x$

After losing 5 marbles,

Number of John’s marbles $=x-5$

Number of Jivanti’s marbles $=45-x-5=40-x$

It is given that the product of their marbles is 124 .

$\therefore(x-5)(40-x)=124$

$\Rightarrow x^{2}-45 x+324=0$

$\Rightarrow x^{2}-36 x-9 x+324=0$

$\Rightarrow x(x-36)-9(x-36)=0$

$\Rightarrow(x-36)(x-9)=0$

Either $x-36=0$ or $x-9=0$

i.e., $x=36$ or $x=9$

If the number of John’s marbles $=36$,

Then, number of Jivanti’s marbles $=45-36=9$

If number of John’s marbles $=9$,

Then, number of Jivanti’s marbles $=45-9=36$

(ii) Let the number of toys produced be $x$.

$\therefore$ Cost of production of each toy $=Rs(55-x)$

It is given that, total production of the toys $=Rs 750$

$\therefore x(55-x)=750$

$\Rightarrow x^{2}-55 x+750=0$

$\Rightarrow x^{2}-25 x-30 x+750=0$

$\Rightarrow x(x-25)-30(x-25)=0$

$\Rightarrow(x-25)(x-30)=0$

Either $x-25=0$ or $x-30=0$

i.e., $x=25$ or $x=30$

Hence, the number of toys will be either 25 or 30 .

Solution

Let the first number be $x$ and the second number is $27-x$.

Therefore, their product $=x(27-x)$

It is given that the product of these numbers is 182 .

Therefore, $x(27-x)=182$

$\Rightarrow x^{2}-27 x+182=0$

$\Rightarrow x^{2}-13 x-14 x+182=0$

$\Rightarrow x(x-13)-14(x-13)=0$

$\Rightarrow(x-13)(x-14)=0$

Either $x-13=0$ or $x-14=0$

i.e., $x=13$ or $x=14$

If first number $=13$, then

Other number $=27-13=14$

If first number $=14$, then

Other number $=27-14=13$

Therefore, the numbers are 13 and 14.

Solution

Let the consecutive positive integers be $x$ and $x+1$.

Given that $x^{2}+(x+1)^{2}=365$

$\Rightarrow x^{2}+x^{2}+1+2 x=365$

$\Rightarrow 2 x^{2}+2 x-364=0$

$\Rightarrow x^{2}+x-182=0$

$\Rightarrow x^{2}+14 x-13 x-182=0$

$\Rightarrow x(x+14)-13(x+14)=0$

$\Rightarrow(x+14)(x-13)=0$

Either $x+14=0$ or $x-13=0$, i.e., $x=-14$ or $x=13$

Since the integers are positive, $x$ can only be 13 .

$\therefore x+1=13+1=14$

Therefore, two consecutive positive integers will be 13 and 14 .

Solution

Let the base of the right triangle be $x cm$.

Its altitude $=(x-7) cm$

From pythagoras theorem,

Base $^{2}+$ Altitude $^{2}=$ Hypotenuse $^{2}$

$\therefore x^{2}+(x-7)^{2}=13^{2}$

$\Rightarrow x^{2}+x^{2}+49-14 x=169$

$\Rightarrow 2 x^{2}-14 x-120=0$

$\Rightarrow x^{2}-7 x-60=0$

$\Rightarrow x^{2}-12 x+5 x-60=0$

$\Rightarrow x(x-12)+5(x-12)=0$

$\Rightarrow(x-12)(x+5)=0$

Either $x-12=0$ or $x+5=0$, i.e., $x=12$ or $x=-5$

Since sides are positive, $x$ can only be 12 .

Therefore, the base of the given triangle is $12 cm$ and the altitude of this triangle will be $(12-7) cm=5 cm$.

Solution

Let the number of articles produced be $x$.

Therefore, cost of production of each article $=Rs(2 x+3)$

It is given that the total production is Rs 90 .

$\therefore x(2 x+3)=90$

$\Rightarrow 2 x^{2}+3 x-90=0$

$\Rightarrow 2 x^{2}+15 x-12 x-90=0$

$\Rightarrow x(2 x+15)-6(2 x+15)=0$

$\Rightarrow(2 x+15)(x-6)=0$

Either $2 x+15=0$ or $x-6=0$, i.e., $x=\dfrac{-15}{2}$ or $x=6$

As the number of articles produced can only be a positive integer, therefore, $x$ can only be 6 .

Hence, number of articles produced $=6$

Cost of each article $=2 \times 6+3=Rs 15$