To Find:
Which of the following quadratic equations has real and equal roots?
Concept:
A quadratic equation has real and equal roots if its discriminant \(D = b^2 - 4ac = 0\)
Let's check each option: Option B:
\[ (x + 1)^2 = 2x + 1 \Rightarrow x^2 + 2x + 1 = 2x + 1 \Rightarrow x^2 + 2x + 1 - 2x - 1 = 0 \Rightarrow x^2 = 0 \Rightarrow \text{Only one root: } x = 0 \Rightarrow \text{Real and equal} \]
✅ This equation has real and equal roots.
Option A:
\[ x^2 + x = 0 \Rightarrow x(x + 1) = 0 \Rightarrow x = 0, x = -1 \Rightarrow \text{Two distinct real roots} \]
❌ Not equal
Option C:
\[ x^2 - 4 = 0 \Rightarrow x = \pm 2 \Rightarrow \text{Two distinct real roots} \]
❌ Not equal
Option D:
\[ x^2 + x + 1 = 0 \Rightarrow D = 1^2 - 4(1)(1) = 1 - 4 = -3 \Rightarrow \text{Imaginary roots} \]
❌ Not real
Final Answer:
✅ The correct option is B. \((x + 1)^2 = 2x + 1\)
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
If the given figure shows the graph of polynomial \( y = ax^2 + bx + c \), then:
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