# Find the values of k for which the given equation has real roots $5{x^2} - kx + 1 = 0$.

Last updated date: 15th Mar 2023

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Answer

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Hint- Here, we will proceed by using the discriminant method for quadratic equations.

Given quadratic equation is $5{x^2} - kx + 1 = 0{\text{ }} \to {\text{(1)}}$

Since we know that for any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$, the roots will be real only if the value of the discriminant is greater than or equal to zero.

On comparing equations (1) and (2), we get

For the given quadratic equation, $a = 5$, $b = - k$ and $c = 1$

Discriminant of any quadratic equation is given by $d = {b^2} - 4ac$

Using above formula, discriminant of the given quadratic equation is given by

\[

d = {\left( { - k} \right)^2} - 4 \times 5 \times 1 \\

\Rightarrow d = {k^2} - 20 \\

\]

Now for the given quadratic equation to have real roots, put $d \geqslant 0$

$ \Rightarrow d \geqslant 0 \Rightarrow {k^2} - 20 \geqslant 0 \Rightarrow {k^2} - {\left( {\sqrt {20} } \right)^2} \geqslant 0 \Rightarrow \left( {k - \sqrt {20} } \right)\left( {k + \sqrt {20} } \right) \geqslant 0$

Clearly for the above inequality to hold true either both the terms on the LHS of the inequality should be positive or both terms should be negative in order to have the product of these terms as positive (greater than or equal to zero)

i.e., either $\left( {k - \sqrt {20} } \right) \geqslant 0 \Rightarrow k \geqslant \sqrt {20} $and $\left( {k + \sqrt {20} } \right) \geqslant 0 \Rightarrow k \geqslant - \sqrt {20} $

or $\left( {k - \sqrt {20} } \right) \leqslant 0 \Rightarrow k \leqslant \sqrt {20} $ and $\left( {k + \sqrt {20} } \right) \leqslant 0 \Rightarrow k \leqslant - \sqrt {20} $

i.e., $k \geqslant \sqrt {20} $ or $k \leqslant - \sqrt {20} $

Since, in case of or, union is performed and in case of and, intersection is performed.

Therefore, The values of $k$ for which the given quadratic equation has real roots are

$k \geqslant \sqrt {20} $ or $k \leqslant - \sqrt {20} $

Note- In these types of problems, we have to compare the given quadratic equation with the general form of any quadratic equation and then in order to have real roots, put the discriminant is greater than or equal to zero.

Given quadratic equation is $5{x^2} - kx + 1 = 0{\text{ }} \to {\text{(1)}}$

Since we know that for any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$, the roots will be real only if the value of the discriminant is greater than or equal to zero.

On comparing equations (1) and (2), we get

For the given quadratic equation, $a = 5$, $b = - k$ and $c = 1$

Discriminant of any quadratic equation is given by $d = {b^2} - 4ac$

Using above formula, discriminant of the given quadratic equation is given by

\[

d = {\left( { - k} \right)^2} - 4 \times 5 \times 1 \\

\Rightarrow d = {k^2} - 20 \\

\]

Now for the given quadratic equation to have real roots, put $d \geqslant 0$

$ \Rightarrow d \geqslant 0 \Rightarrow {k^2} - 20 \geqslant 0 \Rightarrow {k^2} - {\left( {\sqrt {20} } \right)^2} \geqslant 0 \Rightarrow \left( {k - \sqrt {20} } \right)\left( {k + \sqrt {20} } \right) \geqslant 0$

Clearly for the above inequality to hold true either both the terms on the LHS of the inequality should be positive or both terms should be negative in order to have the product of these terms as positive (greater than or equal to zero)

i.e., either $\left( {k - \sqrt {20} } \right) \geqslant 0 \Rightarrow k \geqslant \sqrt {20} $and $\left( {k + \sqrt {20} } \right) \geqslant 0 \Rightarrow k \geqslant - \sqrt {20} $

or $\left( {k - \sqrt {20} } \right) \leqslant 0 \Rightarrow k \leqslant \sqrt {20} $ and $\left( {k + \sqrt {20} } \right) \leqslant 0 \Rightarrow k \leqslant - \sqrt {20} $

i.e., $k \geqslant \sqrt {20} $ or $k \leqslant - \sqrt {20} $

Since, in case of or, union is performed and in case of and, intersection is performed.

Therefore, The values of $k$ for which the given quadratic equation has real roots are

$k \geqslant \sqrt {20} $ or $k \leqslant - \sqrt {20} $

Note- In these types of problems, we have to compare the given quadratic equation with the general form of any quadratic equation and then in order to have real roots, put the discriminant is greater than or equal to zero.

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