Question

If one root of the equation ${{x}^{2}}+px+q=0$ is 2 + $\sqrt{3}$, then the values of p and q are
( a ) -4,1
( b ) 4,-1
( c ) $2\sqrt{3}$
( d ) $-2-\sqrt{3}$

Answer 1

Hint: In this question, a quadratic equation with one of its root is given. First, we compare the equation with the standard quadratic equation, and then by finding the sum of roots and the product of roots and equating the values, we are able to find the value of p and q.

Formula used:
Sum of roots = $-\dfrac{b}{a}$
And product of roots = $\dfrac{c}{a}$

Complete step by step Solution:
Given quadratic equation is ${{x}^{2}}+px+q=0$………………… ( 1)
Compare the equation (1) with the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
a = 1, b = p and c = q
One root of the equation ( 1) is 2 + $\sqrt{3}$
If p and q are rational, then unique values of p and q exist.
If p and q are not rational, then p and q may have infinite values as the other root may or may not be irrational.
First, we assume that p and q are rational numbers
Then the other root of equation (1) must be 2 - $\sqrt{3}$
Now the sum of roots = $-\dfrac{b}{a}$= $-\dfrac{p}{1}$= -p
That is -p = 2 + $\sqrt{3}$ + 2 - $\sqrt{3}$
 Then the value of p =- 4
Now the product of roots = $\dfrac{c}{a}$=$\dfrac{q}{1}$=q
That is ( 2 + $\sqrt{3}$)( 2 - $\sqrt{3}$) = q
Hence the value of q = 1
Hence the value of p = -4 and q = 1

Therefore, the correct option is (a).

Note:We also find out the sum and the product of roots by simply putting the formula if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.