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Show that the equation 3x^2 + 7x + 8 = 0 is not true for any real value of x. [Hint: D = (49 – 4 × 3 × 8) = (49 – 96) = – 47 < 0.]

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Question

Show that the equation 3x2 + 7x + 8 = 0 is not true for any real value of x.

[Hint: D = (49 – 4 × 3 × 8) = (49 – 96) = – 47 < 0.]

Sum
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Solution

Given: 3x2 + 7x + 8 = 0

Step-wise calculation:

1. Compare with ax2 + bx + c = 0:

a = 3, b = 7, c = 8

2. Discriminant D = b2 − 4ac

= 72 – 4 × 3 × 8

= 49 – 96

= –47

3. D < 0, so the quadratic has no real roots (the roots are imaginary).

Since the discriminant is negative, 3x2 + 7x + 8 = 0 has no real solution; the equation is not true for any real value of x.

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Chapter 5: Quadratic Equation - EXERCISE 5C [Page 61]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 5 Quadratic Equation
EXERCISE 5C | Q 22. | Page 61
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