Question

If one root of the quadratic equation ${{x}^{2}}-10x+2k=0$ is $5-\sqrt{3}$ , find k.

Answer 1

Hint: According to the factor theorem if ‘a’ is the root of f(x) then f(a)=0. Use this information to solve this quadratic equation.

Complete step-by-step answer:
The quadratic equation is ${{x}^{2}}-10x+2k=0$.
Value of the given root is $5-\sqrt{3}$.
So according to the factor theorem , the value of the equation at $x=5-\sqrt{3}$ is zero.
So we can write
$\Rightarrow {{\left( 5-\sqrt{3} \right)}^{2}}-10\left( 5-\sqrt{3} \right)+2k=0$
We can expand ${{\left( 5-\sqrt{3} \right)}^{2}}$ by using ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
 $\Rightarrow {{\left( 5 \right)}^{2}}-2\times 5\times \sqrt{3}+{{\left( \sqrt{3} \right)}^{2}}-10\left( 5-\sqrt{3} \right)+2k=0$
$\Rightarrow 25-10\sqrt{3}+3-50+10\sqrt{3}+2k=0$
$\Rightarrow -22+2k=0$
$\Rightarrow 2k=22$
$\Rightarrow k=\dfrac{22}{2}$
$\Rightarrow k=11$
Hence the value of k is 11.

Note: In general the root of any equation is the solution of that equation. For example if we have the equation $x+5=0$. When we solve we will get $x=-5$.
So we can say -5 is the root of the equation .