Mathematics
If -5 is a root of the quadratic equation 2x2 + px - 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then the value of k is:
Quadratic Equations
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Answer
Given,
-5 is a root of the quadratic equation 2x2 + px - 15 = 0.
Substituting value of x = -5 in 2x2 + px - 15 = 0, we get:
⇒ 2(-5)2 + p(-5) - 15 = 0
⇒ 2 × 25 - 5p - 15 = 0
⇒ 50 - 5p - 15 = 0
⇒ 35 - 5p = 0
⇒ 5p = 35
⇒ p =
⇒ p = 7.
Substituting value of p in p(x2 + x) + k = 0, we get:
⇒ 7(x2 + x) + k = 0
⇒ 7x2 + 7x + k = 0
Comparing 7x2 + 7x + k = 0 with ax2 + bx + c = 0 we get,
a = 7, b = 7 and c = k.
Since equation has equal roots,
⇒ Discriminant = 0
⇒ b2 - 4ac = 0
⇒ (7)2 - 4(7)(k) = 0
⇒ 49 - 28k = 0
⇒ 28k = 49
⇒ k =
⇒ k = .
Hence, option 1 is the correct option.
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Case Study I
Shridharacharya was an Indian mathematician, Sanskrit Pandit and philosopher from Bengal. He is known for his treatises – Trisatika and Patiganita. He was the first to give an algorithm for solving quadratic equations. His other major works were on algebra, particularly fractions and he gave an exposition on zero. He separated Algebra from Arithmetic. The quadratic formula which is used to find the roots of a quadratic equation is known as Shridharacharya’s rule.
1. A quadratic equation of the form ax2 + bx + c = 0, a ≠ 0, has two roots given by:
2. A quadratic equation of the form ax2 + bx + c = 0, a ≠ 0, has rational roots, if the value of (b2 - 4ac) is:
- less than 0
- greater than 0
- equal to 0
- equal to 0 or a perfect square
3. The maximum number of roots that a quadratic equation can have is:
- 1
- 2
- 3
- 4
4. A quadratic equation ax2 + bx + c = 0, a ≠ 0, having real coefficients, cannot have real roots if:
- b2 - 4ac < 0
- b2 - 4ac = 0
- b2 - 4ac > 0
- b2 - 4ac ≥ 0
5. The quadratic equation, x2 + 26x + 169 = 0, has:
- non real roots
- rational and unequal roots
- equal roots
- irrational roots
Case Study II
Raman Lal runs a stationery shop in Pune. The analysis of his sales, expenditures and profits showed that for x number of notebooks sold, the weekly profit (in ₹) was P(x) = - 2x2 + 88x - 680. Raman Lal found that:
- He has a loss if he does not sell any notebook in a week.
- There is no profit no loss for a certain value x0 of x.
- The profit goes on increasing with an increase in x i.e. the number of notebooks sold. But he gets a maximum profit at a sale of 22 notebooks in a week.
Now answer the following questions :
1. What will be Raman Lal’s profit if he sold 20 notebooks in a week?
- ₹ 144
- ₹ 280
- ₹ 340
- ₹ 560
2. What is the maximum profit that Raman Lal can earn in a week?
- ₹ 144
- ₹ 288
- ₹ 340
- ₹ 680
3. What is Raman Lal’s loss if he does not sell any notebooks in a particular week?
- ₹ 0
- ₹ 340
- ₹ 680
- ₹ 960
4. Write a quadratic equation for the condition when Raman Lal does not have any profit or loss during a week.
- 2x2 - 44x + 340 = 0
- x2 + 44x - 340 = 0
- x2 - 88x + 340 = 0
- x2 - 44x + 340 = 0
5. What is the minimum number of notebooks x0 that Raman Lal should sell in a week so that he does not incur any loss?
- 0
- 10
- 11
- 12