To determine which quadratic equation does not have real roots, we need to examine the discriminant ($D$) for each equation of the form $ax^2 + bx + c = 0$. The discriminant is calculated using the formula:
$D = b^2 - 4ac$
The nature of the roots depends on the value of the discriminant:
- If $D > 0$, there are two distinct real roots.
- If $D = 0$, there is exactly one real root (a repeated root).
- If $D < 0$, there are no real roots (the roots are complex).
We will calculate the discriminant for each given quadratic equation:
Calculating Discriminant for Quadratic Equations
| Equation | a | b | c | Discriminant ($D = b^2 - 4ac$) | Nature of Roots |
|---|---|---|---|---|---|
| $x^2 + 4x + 4 = 0$ | 1 | 4 | 4 | $D = 4^2 - 4(1)(4) = 16 - 16 = 0$ | One real root |
| $x^2 + 4x + 5 = 0$ | 1 | 4 | 5 | $D = 4^2 - 4(1)(5) = 16 - 20 = -4$ | No real roots |
| $x^2 + 4x - 4 = 0$ | 1 | 4 | -4 | $D = 4^2 - 4(1)(-4) = 16 + 16 = 32$ | Two distinct real roots |
| $x^2 + 4x - 5 = 0$ | 1 | 4 | -5 | $D = 4^2 - 4(1)(-5) = 16 + 20 = 36$ | Two distinct real roots |
Identifying Equation Without Real Roots
Based on the calculations, the equation $x^2 + 4x + 5 = 0$ has a discriminant $D = -4$. Since the discriminant is negative ($D < 0$), this equation does not have real roots.
Similar Questions
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- The equation whose roots are $-2$ and $3$ is:
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- Solve the following :
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Important Questions from Quadratic Equation
- If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of \(\rm \left( 1+\frac{1}{x} \right) \) is equal to :
If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?
Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is
The value of m for which one of the root of x 2 - 3x + 2m = 0 is double of the root of x 2 - x + m = 0 is:
If x 2+ 1 = 2x, then find x – \((\frac{1}{x})\)
