In the quadratic eqn kx2-6x-1=0 determine the values of k for which the eqn does not have any real root
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kx2-6x-1=0
for not having any real root d<0..
hence value of k should be smaller than 0
anandthegreat:
please markk as brainliest
Wrong
Correct, but k is not less than 0, but D less than 0. So, on Solving k < -9
ok!!thanks for correcting me..
Hmm, ok:-)
Answered by
3
Let's consider the general form of differential equation be:
ax² + bx + c = 0
As we know, Discriminant (D) = b² - 4ac
Important:
★ For Real Roots, D ≥ 0
★For Imaginary_(Non real) roots, D < 0
★ For real & EQUAL roots, D = 0
★ For real & UNEQUAL roots, D < 0
SO, according to Question:
Quadratic Equation is kx² - 6x - 1 = 0
For No real roots, "D < 0", So,
Discriminant (D) = (-6)² - 4(k)(-1) < 0
D = 36 + 4k < 0
→ 4k < -36
→ k < -9
So, for all "k" Less than -9", This quadratic equation does not have any real roots.
Thankyou!!!
ax² + bx + c = 0
As we know, Discriminant (D) = b² - 4ac
Important:
★ For Real Roots, D ≥ 0
★For Imaginary_(Non real) roots, D < 0
★ For real & EQUAL roots, D = 0
★ For real & UNEQUAL roots, D < 0
SO, according to Question:
Quadratic Equation is kx² - 6x - 1 = 0
For No real roots, "D < 0", So,
Discriminant (D) = (-6)² - 4(k)(-1) < 0
D = 36 + 4k < 0
→ 4k < -36
→ k < -9
So, for all "k" Less than -9", This quadratic equation does not have any real roots.
Thankyou!!!
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