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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Important Questions from Quadratic Equation
If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\) is:
If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:
If x 2 – 3x + 1 = 0, then the value of \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\) is:
If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) , \(x \ne 0\) , then what is the value of \((x^4+{1\over{x^2}})\over(x^2+1) \) ?
If x 2 + \(\frac{1}{x^2}\) = 18, x > 0, then find the value of x 3 + \(\frac{1}{x^3}\) .
