Question

Show that the equation ${x^2} + 2px - 3 = 0$ has real and distinct roots for all values of p.

Answer 1

We are given with an equation and need to find the nature of its roots. We know that nature of the roots is determined from its discriminant of the form $D = {b^2} - 4ac$ . The nature of the roots is
1.If ${b^2} - 4ac < 0$ then the equation has no real root.
2.If ${b^2} - 4ac = 0$ then the equation has equal and real roots.
3.If ${b^2} - 4ac > 0$ then the equation has unequal but real roots.
From these conditions we can determine the nature of the roots of the above equation

Complete step-by-step answer:
Given that,
${x^2} + 2px - 3 = 0$
Comparing this with the general form of quadratic equation we get, a=1, b=2p and c=-3.
Thus discriminant D is given by,
$D = {b^2} - 4ac$
$\Rightarrow {\left( {2p} \right)^2} - 4 \times 1 \times \left( { - 3} \right)$
$\Rightarrow {\left( {2p} \right)^2} + 12$
This is the value of D and it shows that this value is always greater than zero.
$D = {\left( {2p} \right)^2} + 12 > 0$
From the nature of the roots we clearly get that this equation has real and distinct roots for all values of p.

Note: We are not going to check for any particular value of p here but in general we can say the statement of nature of roots. Students also need not to find any particular value of p and then that of roots. Only check the nature of discriminant.