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For what value of k are the roots of the quadratic equation k x(x-2 \sqrt{5})+10=0 real and equal.
CBSE | Class 10 | Exercise 10D | Maths | Quadratic Equations | RS Aggarwal
For what value of k are the roots of the quadratic equation k x(x-2 \sqrt{5})+10=0 real and equal.

The given equation is

\begin{array}{l} k x(x-2 \sqrt{5})+10=0 \\ \Rightarrow k x^{2}-2 \sqrt{5} k x+10=0 \end{array}

This is of the form a x^{2}+b x+c=0, where a=k, b=-2 \sqrt{5} k and
c=10.

\therefore D=b^{2}-4 a c=(-2 \sqrt{5} k)^{2}-4 \times k \times 10=20 k^{2}-40 k

The given equation will have real and equal roots if D=0.
\begin{array}{l} \therefore 20 k^{2}-40 k=0 \\ \Rightarrow 20 k(k-2)=0 \\ \Rightarrow k=0 \text { or } k-2=0 \\ \Rightarrow k=0 \text { or } k=2 \end{array}

But, for k=0, we get 10=0, which is not true

Hence, 2 is the required value of k.

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