prove that the roots of x²+(1-k)x+k-3=0 are real for all real values of k
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EXPLANATION.
Roots of quadratic equation.
⇒ x² + (1 - k)x + k - 3 = 0.
As we know that,
⇒ D = Discriminant or b² - 4ac.
⇒ D = (1 - k)² - 4(1)(k - 3).
As we know that,
Formula of :
⇒ (x - y)² = x² + y² - 2xy.
Using this formula in equation, we get.
⇒ D = (1 + k² - 2k) - 4(k - 3).
⇒ D = 1 + k² - 2k - 4k + 12.
⇒ D = k² - 6k + 13.
Again we factorizes the equation.
⇒ D = (-6)² - 4(1)(13).
⇒ D = 36 - 52.
⇒ D = -16.
For real roots, D ≥ 0.
And the value of this equation is D < 0 it means roots are imaginary.
There is no real values exists in the equation.
MORE INFORMATION.
Nature of the roots 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.
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