find the value of k for which equations x²-4x+k=0 have real and equal root
Answers
Answer:
k=4
Step-by-step explanation:
For real and equal roots D=0
b^2-4ac=0
-4^2-4×1×k=0
16-4k=0
16=4k
k=16÷4
k=4
Answer:
The value of k is 4.
Step-by-step-explanation:
The given quadratic equation is
x² - 4x + k = 0.
Comparing with ax² + bx + c = 0, we get,
- a = 1
- b = - 4
- c = k
For real and equal roots,
b² - 4ac = 0
⇒ ( - 4 )² - 4 * 1 * k = 0
⇒ 16 - 4k = 0
⇒ 16 = 4k
⇒ k = 16 / 4
⇒ k = 4
∴ The value of k is 4.
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Additional Information:
1. Quadratic Equation:
An equation with degree 2 is called a quadratic equation.
The general form of quadratic equation is ax² + bx + c = 0, where a ≠ 0 and a, b, c are real numbers.
A quadratic equation has two roots, i. e. two values of x which satisfy the equation.
2. Nature of roots:
The nature of roots of quadratic equation depends upon the value of discriminant.
The discriminant of quadratic equation is b² - 4ac.
1. If b² - 4ac > 0, the roots are real and unequal.
2. If b² - 4ac = 0, the roots are real and equal.
3. If b² - 4ac < 0, the roots are imaginary.
4. If b² - 4ac ≥ 0, the roots are real.