Answer :
Sure, lets go step-by-step:
### (a) Making a Quadratic Equation for the Above Situation
Let's denote the original price of a book by [tex]\( x \)[/tex] rupees.
- According to the problem, if the price per book is increased by Rs. 10, the person can buy 10 fewer books for Rs. 1200.
- The total amount of Rs. 1200 can be expressed as the number of books bought multiplied by the price per book.
Let's establish our equation:
1. Number of books bought originally: [tex]\( \frac{1200}{x} \)[/tex]
2. New price per book: [tex]\( x + 10 \)[/tex]
3. Number of books that could be bought at the increased price: [tex]\( \frac{1200}{x + 10} = \frac{1200}{x + 10} = \frac{1200}{x} - 10 \)[/tex]
Using these facts, we can express the situation with one equation:
[tex]\[ \left(\frac{1200}{x} - 10\right) \times (x + 10) = 1200 \][/tex]
Let's simplify this equation to form a quadratic equation:
[tex]\[ \left(\frac{1200 - 10x}{x}\right) \times (x + 10) = 1200 \][/tex]
[tex]\[ (1200 - 10x) \times (1 + \frac{10}{x}) = 1200 \][/tex]
[tex]\[ (1200 - 10x) \left(\frac{x + 10}{x}\right) = 1200 \][/tex]
Multiplying both sides by [tex]\( x \)[/tex]:
[tex]\[ (1200 - 10x)(x + 10) = 1200x \][/tex]
Expanding gives:
[tex]\[ 1200x + 12000 - 10x^2 - 100x = 1200x \][/tex]
Rearranging terms, we obtain:
[tex]\[ -10x^2 + 11000 = 0 \][/tex]
Dividing the equation by -10:
[tex]\[ x^2 + 10x - 1200 = 0 \][/tex]
Thus, the quadratic equation is:
[tex]\[ x^2 + 10x - 1200 = 0 \][/tex]
### (b) Finding the Original List Price of the Book
Now let's solve the quadratic equation [tex]\( x^2 + 10x - 1200 = 0 \)[/tex].
The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here,
[tex]\[ a = 1, b = 10, c = -1200 \][/tex]
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-1200)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{100 + 4800}}{2} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{4900}}{2} \][/tex]
[tex]\[ x = \frac{-10 \pm 70}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{60}{2} = 30 \][/tex]
[tex]\[ x = \frac{-80}{2} = -40 \][/tex]
### (c) Is there a negative value of [tex]\( x \)[/tex]? If so, why do you not accept this value? Give the reason.
Yes, there is a negative value of [tex]\( x \)[/tex], which is -40. However, we do not accept this value because the price of a book cannot be negative in a real-world scenario. Hence the valid solution to the original list price of the book is:
[tex]\[ x = 30 \text{ rupees} \][/tex]
### (a) Making a Quadratic Equation for the Above Situation
Let's denote the original price of a book by [tex]\( x \)[/tex] rupees.
- According to the problem, if the price per book is increased by Rs. 10, the person can buy 10 fewer books for Rs. 1200.
- The total amount of Rs. 1200 can be expressed as the number of books bought multiplied by the price per book.
Let's establish our equation:
1. Number of books bought originally: [tex]\( \frac{1200}{x} \)[/tex]
2. New price per book: [tex]\( x + 10 \)[/tex]
3. Number of books that could be bought at the increased price: [tex]\( \frac{1200}{x + 10} = \frac{1200}{x + 10} = \frac{1200}{x} - 10 \)[/tex]
Using these facts, we can express the situation with one equation:
[tex]\[ \left(\frac{1200}{x} - 10\right) \times (x + 10) = 1200 \][/tex]
Let's simplify this equation to form a quadratic equation:
[tex]\[ \left(\frac{1200 - 10x}{x}\right) \times (x + 10) = 1200 \][/tex]
[tex]\[ (1200 - 10x) \times (1 + \frac{10}{x}) = 1200 \][/tex]
[tex]\[ (1200 - 10x) \left(\frac{x + 10}{x}\right) = 1200 \][/tex]
Multiplying both sides by [tex]\( x \)[/tex]:
[tex]\[ (1200 - 10x)(x + 10) = 1200x \][/tex]
Expanding gives:
[tex]\[ 1200x + 12000 - 10x^2 - 100x = 1200x \][/tex]
Rearranging terms, we obtain:
[tex]\[ -10x^2 + 11000 = 0 \][/tex]
Dividing the equation by -10:
[tex]\[ x^2 + 10x - 1200 = 0 \][/tex]
Thus, the quadratic equation is:
[tex]\[ x^2 + 10x - 1200 = 0 \][/tex]
### (b) Finding the Original List Price of the Book
Now let's solve the quadratic equation [tex]\( x^2 + 10x - 1200 = 0 \)[/tex].
The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here,
[tex]\[ a = 1, b = 10, c = -1200 \][/tex]
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-1200)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{100 + 4800}}{2} \][/tex]
[tex]\[ x = \frac{-10 \pm \sqrt{4900}}{2} \][/tex]
[tex]\[ x = \frac{-10 \pm 70}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{60}{2} = 30 \][/tex]
[tex]\[ x = \frac{-80}{2} = -40 \][/tex]
### (c) Is there a negative value of [tex]\( x \)[/tex]? If so, why do you not accept this value? Give the reason.
Yes, there is a negative value of [tex]\( x \)[/tex], which is -40. However, we do not accept this value because the price of a book cannot be negative in a real-world scenario. Hence the valid solution to the original list price of the book is:
[tex]\[ x = 30 \text{ rupees} \][/tex]
