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# The values of k for which the roots are real and equal of the following equation $k{x^2} - 2\sqrt 5 x + 4 = 0$ is $k = \dfrac{5}{4}$ .(a) True (b) False

Last updated date: 02nd Aug 2024
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Hint:For the real and equal roots we know that the discriminant is equal to $0$ or for the equation $a{x^2} + bx + c = 0$ the condition for real and equal is ${b^2} - 4ac = 0$ from this concept we can solve this question easily.

We know that the condition for real and equal roots Discriminant = $0$ .
If equation is $a{x^2} + bx + c = 0$ then Discriminant is = ${b^2} - 4ac$ ;
Now compare the both equation for getting the values of $a,b,c$ :
$k{x^2} - 2\sqrt 5 x + 4 = 0$ with $a{x^2} + bx + c = 0$ ;
By comparing we get
$a = k$ , $b = - 2\sqrt 5$ , $c = 4$
Now Discriminant of required equation is
${b^2} - 4ac$ = ${\left( { - 2\sqrt 5 } \right)^2} - 4 \times k \times 4 = 0$
By solving we get
$20 - 16k = 0$ ,
Now by rearranging we get,
$k = \dfrac{{20}}{{16}}$
Dividing Numerator or denominator by $4$ we get ,
$k = \dfrac{5}{4}$
Hence the value given is correct.

So, the correct answer is “True”.

Note:If in case we have to find the value of $k$ for two real and unequal roots then we have to do that Discriminant $> 0$ or ${b^2} - 4ac$$> 0 .If in case we have to find the value of k for imaginary roots then we have to do that Discriminant < 0 or {b^2} - 4ac$$ < 0$.When the Discriminant is negative then we get the pair of complex equations .If one root is given $a + ib$ than other will be the $a - ib$ .