हिंदी

Discuss the nature of the roots of the following equation without actually solving it: 16x^2 = 24x + 1

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प्रश्न

Discuss the nature of the roots of the following equation without actually solving it:

16x2 = 24x + 1

योग
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उत्तर

Given: 16x2 = 24x + 1

Step-wise calculation:

1. Put in standard form:

16x2 – 24x – 1 = 0

So a = 16, b = –24, c = –1.

2. Discriminant: Δ = b2 – 4ac

= (–24)2 – 4(16)(–1) 

= 576 + 64

= 640

Since Δ > 0, the equation has two distinct real roots.

3. Simplify `sqrt(Δ)`:

`sqrt(640) = sqrt(64 xx 10)`

 = `8sqrt(10)`

Because 10 is not a perfect square, `sqrt(10)` is irrational, so `sqrt(Δ)` is irrational.

4. By the quadratic formula the roots are `(-b ± sqrt(Δ))/(2a)`. 

Here `(-b)/(2a) = 24/32 = 3/4` is rational, while `sqrt(Δ)/(2a) = (8sqrt(10))/(32) = (sqrt(10))/4` is irrational.

Adding or subtracting a nonzero irrational to a rational gives two irrational numbers; the ± gives two different values.

Therefore, the two roots are irrational and unequal.

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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 5. | पृष्ठ ६१
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