# value of $k$, for which both roots of $x^2-8kx+16(k^2-k+1)=0$ are real and Distinct.

The smallest value of $k,$ for which both the roots of the equation

$x^2 – 8kx + 16 (k^2 – k + 1) = 0$ are real, distinct and have values at least $4,$ is

$\bf{My\; Trial::}$ If the Given equation has real and distinct solution, Then $\bf{D(discriminant\geq 0)}$

So $64k^2-16(k^2-k+1)>0\Rightarrow k-1>0\Rightarrow k>1........................(1)$

Now Both Roots have least value is $=4$.

Now Let $\bf{\alpha}$ and $\bf{\beta}$ be the roots of $x^2-8kx+16(k^2-k+1)=0.$ Then $\bf{\alpha+\beta = 8k}$

So Given $\alpha+\beta >8\;\;\;$(bcz Roots are Different.)

So $8k>8\Rightarrow k>1.................................................................................(2)$

Now answer Given is $k=2$

But I did not understand How can I calculate it.

Help me

Thanks.

• Careful. If you want real and distinct roots, then the discriminant must be strictly greater than $0$. – Kaj Hansen Jun 21 '14 at 4:37

Your equation can be rewritten as $$(x-4k)^2=16(k-1).$$ This simplifies to $$x=4\left(k \pm \sqrt{k-1}\right)$$ For the roots to be distinct and $\geq 4$ we want $$k > 1 \qquad \text{and } \qquad k \pm\sqrt{k-1} \geq 1$$ As you can see there are infinitely many values of $k$ that can satisfy the above inequalities but the smallest value will be $k=2$ (this will come from having $k-\sqrt{k-1} \geq 1$.