हिंदी

Show that the equation x^2 + ax – 1 = 0 has real and distinct roots for all real values of x. [Hint: D = (a^2 + 4) > 0.]

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प्रश्न

Show that the equation x2 + ax – 1 = 0 has real and distinct roots for all real values of x.

[Hint: D = (a2 + 4) > 0.]

योग
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उत्तर

Given: Consider the quadratic equation x2 + ax – 1 = 0

Step-wise calculation:

1. Compare with the standard form Ax2 + Bx + C = 0: 

A = 1, B = a, C = –1

2. Compute the discriminant D = B2 – 4AC

= a2 – 4(1)(–1)

= a2 + 4

3. For every real a, a2 ≥ 0, so a2 + 4 ≥ 4 > 0.

Hence, D = a2 + 4 > 0 for all real a.

4. If the discriminant D > 0, a quadratic has two real and distinct roots.

Since D = a2 + 4 > 0 for every real a, the equation x2 + ax – 1 = 0 has two real and distinct roots for all real values of a.

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अध्याय 5: Quadratic Equation - EXERCISE 5C [पृष्ठ ६१]

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आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 5 Quadratic Equation
EXERCISE 5C | Q 21. | पृष्ठ ६१
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