If $x^2-(a-3) x+a=0$ has atleast one positive root, then

$a \in(-\infty, 0) \cup[7, \infty)$

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

x^2-(a-3) x+a=0

has atleast one positive root, then

View answer and explanation

Correct option: C

a \in(-\infty, 0] \cup[9, \infty)

Option C (

a \in(-\infty, 0] \cup[9, \infty)

) is correct because it includes all values of 'a' for which the quadratic equation has at least one positive root, considering both cases where one or both roots are positive.

More Questions from QUADRATIC EQUATIONS

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The solution of the equation

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

is If

\alpha ,\beta

are the roots of

{x^2} - 3x + a = 0,a \in R

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

{x^2} - qx + p = 0,

then the equation whose roots are

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

is If

{x^2} + 4ax + 2 > 0

for all values of

x

, then

a

lies in the interval If

f\left( x \right) = a{x^2} + bx + c\;

and

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

and

\;a \ne 0,

then Root(s) of the equation

9{x^2} - 18\left| x \right| + 5 = 0

belonging to the domain of definition of the function

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)

is (are)