(i) $x^{2} – 3x – 10$ = 0
(ii) $2x^{2} + x – 6$ = 0
(iii) $√2 x^{2} + 7x + 5√2$ = 0
(iv) $2x^{2} – x +\dfrac{1}{8}$ = 0
(v) $100x^{2} – 20x + 1$ = 0
(i) Given,$ x^{2} – 3x – 10$ =0
Taking LHS,
=>$x^{2} – 5x + 2x – 10$
=>x(x – 5) + 2(x – 5)
=>(x – 5)(x + 2)
The roots of this equation, $x^{2} – 3x – 10$ = 0 are the values of x for which (x – 5)(x + 2) = 0
Therefore, x – 5 = 0 or x + 2 = 0
=> x = 5 or x = -2
(ii) Given, $2x^{2} + x – 6$ = 0
Taking LHS,
=> $2x^{2} + 4x – 3x – 6$
=> 2x(x + 2) – 3(x + 2)
=> (x + 2)(2x – 3)
The roots of this equation, $2x^{2} + x – 6$=0 are the values of x for which (x + 2)(2x – 3) = 0
Therefore, x + 2 = 0 or 2x – 3 = 0
=> x = -2 or x = $\dfrac{3}{2}$
(iii) $\sqrt{2} x^{2} + 7x + 5\sqrt{2}$=0
Taking LHS,
=> $\sqrt{2} x^{2} + 5x + 2x + 5\sqrt{2}$
=> $x (\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5)$=$ (\sqrt{2}x + 5)(x + \sqrt{2})$
The roots of this equation, $\sqrt{2} x^{2} + 7x + 5\sqrt{2}$=0 are the values of x for which $(\sqrt{2}x + 5)(x + \sqrt{2})$ = 0
Therefore, $\sqrt{2}x + 5 = 0 or x + \sqrt{2}$ = 0
=> x =$ -5/\sqrt{2}$ or x = $-\sqrt{2}$
(iv)$ 2x^{2} – x +\dfrac{1}{8} $= 0
Taking LHS,
=$\dfrac{1}{8} (16x^{2} – 8x + 1)$
= $\dfrac{1}{8} (16x^{2} – 4x -4x + 1)$
= $\dfrac{1}{8} (4x(4x – 1) -1(4x – 1))$
= $\dfrac{1}{8} (4x – 1)^{2} $
The roots of this equation, $2x^{2} – x + \dfrac{1}{8} $= 0, are the values of x for which $(4x – 1)^{2}$ = 0
Therefore, (4x – 1) = 0 or (4x – 1) = 0
⇒ x = $\dfrac{1}{4} $or x = $\dfrac{1}{4}$
(v) Given, $100x^{2} – 20x + 1$=0
Taking LHS,
= $100x^{2} – 10x – 10x + 1$
= 10x(10x – 1) -1(10x – 1)
= $(10x – 1)^{2} $
The roots of this equation,$ 100x^{2} – 20x + 1$=0, are the values of x for which $(10x – 1)^{2}$ = 0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒x = $\dfrac{1}{10}$ or x = $\dfrac{1}{10}$
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.
(i) Let us say, the number of marbles John has = x.
Therefore, number of marbles Jivanti has = 45 – x
After losing 5 marbles each,
Number of marbles John has = x – 5
Number of marbles Jivanti has = 45 – x – 5 = 40 – x
Given that the product of their marbles is 124.
∴ (x – 5)(40 – x) = 124
⇒ $x^{2} – 45x + 324$ = 0
⇒$ x^{2} – 36x – 9x + 324$ = 0
⇒ x(x – 36) -9(x – 36) = 0
⇒ (x – 36)(x – 9) = 0
Thus, we can say,
x – 36 = 0 or x – 9 = 0
⇒ x = 36 or x = 9
Therefore,
If, John’s marbles = 36,
Then, Jivanti’s marbles = 45 – 36 = 9
And if John’s marbles = 9,
Then, Jivanti’s marbles = 45 – 9 = 36
(ii) Let us say, number of toys produced in a day be x.
Therefore, cost of production of each toy = Rs(55 – x)
Given, total cost of production of the toys = Rs 750
∴ x(55 – x) = 750
⇒ $x^{2} – 55x + 750$ = 0
⇒ $x^{2} – 25x – 30x + 750$ = 0
⇒ x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
Thus, either x -25 = 0 or x – 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys produced in a day, will be either 25 or 30.
Let us say, first number be x and the second number is 27 – x.
Therefore, the product of two numbers
x(27 – x) = 182
⇒$ x^{2} – 27x – 182$ = 0
⇒ $x^{2} – 13x – 14x + 182$ = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
Thus, either, x = -13 = 0 or x – 14 = 0
⇒ x = 13 or x = 14
Therefore, if first number = 13, then second number = 27 – 13 = 14
And if first number = 14, then second number = 27 – 14 = 13
Hence, the numbers are 13 and 14.
Let us say, the two consecutive positive integers be x and x + 1.
Therefore, as per the given questions,
$x^{2} + (x + 1)^{2} $= 365
$⇒ x^{2} + x^{2} + 1 + 2x$ = 365
$⇒ 2x^{2} + 2x – 364$ = 0
$⇒ x^{2} + x – 182$ = 0
$⇒ x^{2} + 14x – 13x – 182$ = 0
$⇒ x(x + 14) -13(x + 14)$ = 0
$⇒ (x + 14)(x – 13) $= 0
Thus, either, x + 14 = 0 or x – 13 = 0,
⇒ x = – 14 or x = 13
since, the integers are positive, so x can be 13, only.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.
Let us say, the base of the right triangle is x cm.
Given, the altitude of right triangle = (x – 7) cm
From Pythagoras theorem, we know,
$Base^{2} + Altitude^{2}$= $Hypotenuse^{2} $
∴ $x^{2} + (x – 7)^{2}$ = $13^{2}$
$⇒ x^{2} + x^{2} + 49 – 14x $= 169
$⇒ 2x^{2} – 14x – 120$ = 0
$⇒ x^{2} – 7x – 60$ = 0
$⇒ x^{2} – 12x + 5x – 60 $= 0
$⇒ x(x – 12) + 5(x – 12)$ = 0
$⇒ (x – 12)(x + 5)$ = 0
Thus, either x – 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = – 5
Since sides cannot be negative, 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.
Let us say, the number of articles produced is x.
Therefore, cost of production of each article = Rs. (2x + 3)
Given, total cost of production is Rs. 90
∴ x(2x + 3) = 90
$⇒ 2x^{2} + 3x – 90 $= 0
$⇒ 2x^{2} + 15x -12x – 90$ = 0
⇒ x(2x + 15) -6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
Thus, either 2x + 15 = 0 or x – 6 = 0
⇒ x = $-\dfrac{15}{2} $or x = 6
As the number of articles produced can only be a positive integer, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs. 15.