To solve the question, "Which of the following equations has no real roots?" let's analyze each option step by step. A quadratic equation has no real roots if its discriminant ($\Delta$) is negative. The discriminant is given by:

$\Delta = b^2 - 4ac$

where the quadratic equation is written in standard form: $ax^2 + bx + c = 0$

Analyzing Each Option

(a) $x^2 - 4x + 4 = 0$:

• Here, $a = 1$, $b = -4$, and $c = 4$.
• Discriminant: $\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$
• Since $\Delta = 0$, this equation has real and repeated roots.

(b) $x^2 - 4x = 0$:

• Rewrite as $x^2 - 4x + 0 = 0$. Here, $a = 1$, $b = -4$, and $c = 0$.
• Discriminant: $\Delta = (-4)^2 - 4(1)(0) = 16 - 0 = 16$
• Since $\Delta > 0$, this equation has two distinct real roots.

(c) $3x^2 - 1 = 0$:

• Rewrite as $3x^2 - 1 + 0 = 0$. Here, $a = 3$, $b = 0$, and $c = -1$.
• Discriminant: $\Delta = (0)^2 - 4(3)(-1) = 0 + 12 = 12$
• Since $\Delta > 0$, this equation has two distinct real roots.

(d) $x^2 + 1 = 0$:

• Rewrite as $x^2 + 0x + 1 = 0$. Here, $a = 1$, $b = 0$, and $c = 1$.
• Discriminant: $\Delta = (0)^2 - 4(1)(1) = 0 - 4 = -4$
• Since $\Delta < 0$, this equation has no real roots (the roots are complex).

Final Answer:

The equation with no real roots is (d) $x^2 + 1 = 0$.

Would you like further details or explanations?

Related Questions:

1. What does it mean if the discriminant ($\Delta$) equals 0?
2. How can we find the roots of a quadratic equation if the discriminant is negative?
3. Why does the equation $x^2 + 1 = 0$ result in complex roots?
4. Can a quadratic equation have only one real root? If so, when?
5. How do the coefficients of a quadratic equation affect its graph?

Tip:

For any quadratic equation, always check the discriminant first to determine the nature of the roots (real, repeated, or complex).