Question

If one of the root of x^2+bx+c=0 (where band c are rational numbers) is 2-root3, then |b+c| is equal to?

Please answer with method and don't spam if you don't know.​

Answer 1

Answer:

Let us recall the general solution, α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a

Case I: D^2 = b2 – 4ac > 0

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive, then the roots α and β of the quadratic equation ax2 +bx+ c = 0 are real and unequal.

Case II: b2– 4ac = 0

When a, b, and c are real numbers, a ≠ 0 and discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.

Case III: b2– 4ac < 0

When a, b, and c are real numbers, a ≠ 0 and discriminant is negative, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are unequal and not real. In this case, we say that the roots are imaginary.

Case IV: b2 – 4ac > 0 and perfect square

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real, rational and unequal.

Case V: b2– 4ac > 0 and not perfect square

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax2 + bx + c = 0 are real, irrational and unequal.

Here the roots α and β form a pair of irrational conjugates.

Case VI: b2– 4ac >0 is perfect square and a or b is irrational

When a, b, and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax2 + bx + c = 0 are irrational.

Given question is case no (iv) so option B is true.

Answer 2

Given,

$\[{x^2} + bx + c = 0\]$

Root, $\[=2 - \sqrt 3 \]$

Solution,

Know that quadratic equation has 2 roots. One root is $\[2 - \sqrt 3 \]$ then other roots will be $\[2 + \sqrt 3 \]$

Format of quadratic equation,

$\[{x^2} + \left( {sum\,of\,roots} \right)x + product\,of\,roots = 0\]$

Sum of roots $\[ = b\]$

$\[\begin{array}{l}b = \left( {2 - \sqrt 3 } \right) + \left( {2 + \sqrt 3 } \right)\\ \Rightarrow b = 4\end{array}\]$

Product of roots $\[ = c\]$

$\[\begin{array}{l}c = \left( {2 - \sqrt 3 } \right) \times \left( {2 + \sqrt 3 } \right)\\ \Rightarrow c = \left( {{2^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)\\ \Rightarrow c = \left( {4 - 3} \right)\\ \Rightarrow c = 1\end{array}\]$

Calculate the value of $\[|b + c|\]$,

$\[\begin{array}{l} = |b + c|\\ = |4 + 1|\\ = 5\end{array}\]$

Hence the value is 5.