Question

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Answer 1

Solution :

Case 1

• Sum of zeroes = 0
• Product of zeroes = √5

$\implies\tt {x}^{2} - (sum)x + product \\ \\ \implies\tt {x}^{2} - (0)x + \sqrt{5} \\ \\ \implies \tt {x}^{2} + \sqrt{5}$

Case 2

• Sum of zeroes = - 7/4
• Product of zeroes = 1/4

$\implies \tt {x}^{2} - (sum)x + product \\ \\ \tt \implies {x}^{2} - \bigg( - \frac{7}{4} \bigg)x + \frac{1}{4} \\ \\ \tt \implies {x}^{2} + \frac{7x}{4} + \frac{1}{4}$

Case 3

• Sum of zeroes = 4
• Product of zeroes = 1

$\implies\tt {x}^{2} - (sum)x + product \\ \\ \tt \implies {x}^{2} - (4)x + 1 \\ \\ \tt \implies {x}^{2} - 4x + 1$

Answer 2

Question:

Use the formula :-

p(x) = x^2 - (sum of zero) x + product of zeroes.

Find quadratic polynomial each with given numbers as dum & Product of its zeroes:-

1. 0, √5
2. -7/4, 1/4
3. 4, 1

Solution :

Case 1

• Sum of zeroes = 0
• Product of zeroes = √5

$\implies {x}^{2} - (sum)x + product$

$\implies {x}^{2} - (0)x + \sqrt{5}$

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

Case 2

• Sum of zeroes = - 7/4
• Product of zeroes = 1/4

$\implies {x}^{2} - (sum)x + product$

$\implies {x}^{2} - \frac({-7}{4}x + \frac{1}{4}$

$\implies {x}^{2} + \frac{7x}{4}x + \frac{1}{4}$

Case 3

• Sum of zeroes = 4
• Product of zeroes = 1

$\implies {x}^{2} - (sum)x + product$

$\implies {x}^{2} - (4)x + 1$

$\implies {x}^{2} - 4x + 1$