Q.IV. 1) Find C if the quadratic equation x^2-2(C+1)x +c^2=0 has real and
equal roots.
Answers
EXPLANATION.
Quadratic equation,
⇒ F(x) = x² - 2(c + 1)x + c² = 0.
As we know that,
For real and equal roots,
D = 0 Or b² - 4ac = 0.
⇒ [-2(c + 1)²] - 4(1)(c²) = 0.
⇒ [4(c² + 1 + 2c] - 4c² = 0.
⇒ [4c² + 4 + 8c] - 4c² = 0.
⇒ 4c² + 4 + 8c - 4c² = 0.
⇒ 8c + 4 = 0.
⇒ 8c = -4.
⇒ c = -1/2.
MORE INFORMATION.
Nature of the factors of the quadratic equation,
(1) = Real and different, if b² - 4ac > 0.
(2) = Rational and different, if b² - 4ac is a perfect square.
(3) = Real and equal, if b² - 4ac = 0.
(4) = if D < 0, Roots are imaginary and unequal or complex conjugate.
Answer:
-1/2
Step-by-step explanation:
the one and only condition for the roots to be real and equal is determinant of the equation should be zero.
in the equation ax*2 +bx+c the determinant is b*2- 4ac .
so in the above question determinant is
4(c+1)*2- 4c*2 so the condition is it should be zero.
4c*2+8c+4-4c*2=0
8c+4=0
c= -4/8 = -1/2