Question

Find the value of k for which the given quadratic equation \[{{x}^{2}}-4x+k=0\] distinct real roots.

Answer 1

Hint : Whenever we have the nature of roots given in question we use discriminant of quadratic equations. We can find discriminant by using below formula
\[d={{b}^{2}}-4ac\]
Where d – discriminant
b - Coefficient of \['x'\] in polynomial
a – coefficient of\['{{x}^{2}}'\] in polynomial
c – constant term of polynomial
Nature of Discriminant –
\[d > 0\text{ ,2}\] real & distinct roots
\[d=0\text{ },1\] real roots (equal)
\[d < 0\text{ ,2}\] imaginary roots
Complete step-by-step answer:
Given, \[{{x}^{2}}-4x+k=0\]
On comparing with \[\text{a}{{\text{x}}^{2}}\text{+bx+c}\]
\[\text{a =1, b = }-4,\text{ }c=k\]
According to question, distinct real roots i.e., \[d\ge 0\]
\[\Rightarrow {{b}^{2}}-4\times 1\times k\ge 0\]
\[\Rightarrow 16-4k\ge 0\]
\[\Rightarrow -4k\ge -16\]
\[\Rightarrow 4k\le 16\]
\[\Rightarrow k\le 4\]
Note: Discriminants can be positive or negative.
While doing this question we need to be careful about inequality signs. When we multiply or divide both sides of inequality by negative sign then inequality sign gets reversed means greater than changes to less than and less than to greater than.