Question

The positive root of the equation $\left( {\sqrt {200} + \sqrt {56} } \right){x^2} + 10x - 2\left( {\sqrt {50} - \sqrt {14} } \right) = 0$
A) $\dfrac{{\sqrt {26} }}{{\sqrt {14} }}$
B) $\sqrt {200} - \sqrt {56} $
C) $\dfrac{{5\sqrt 2 - 14}}{9}$
D) $\dfrac{{10}}{{\sqrt {200} - \sqrt {56} }}$

Answer 1

Hint: First we will compare the given quadratic equation with the general quadratic equation $a{x^2} + bx + c = 0$ to get the values of a, b, c. Then we will apply the quadratic formula for a quadratic equation $a{x^2} + bx + c = 0$ i.e. \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to get both the roots. Out of the two roots, we will select the positive root as our answer.

Complete step by step solution:
For a quadratic equation, $a{x^2} + bx + c = 0$, by quadratic formula, we have:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Comparing, the above equation $\left( {\sqrt {200} + \sqrt {56} } \right){x^2} + 10x - 2\left( {\sqrt {50} - \sqrt {14} } \right) = 0$with $a{x^2} + bx + c = 0$
We can see that, $a = \sqrt {200} + \sqrt {56} $
$b = 10$ and $c = - 2\left( {\sqrt {50} - \sqrt {14} } \right)$
Now, \[{b^2} = {\left( {10} \right)^2} = 100\]
And, \[4ac = 4\left( {\sqrt {200} + \sqrt {56} } \right)\left( { - 2\left( {\sqrt {50} - \sqrt {14} } \right)} \right)\]
$\Rightarrow $4ac = - 8\left( {\sqrt {200} + \sqrt {56} } \right)\left( {\sqrt {50} - \sqrt {14} } \right)\]
$\Rightarrow $\[4ac = - 8\left( {2\sqrt {50} + 2\sqrt {14} } \right)\left( {\sqrt {50} - \sqrt {14} } \right)\]
$\Rightarrow $\[4ac = - 16\left( {\sqrt {50} + \sqrt {14} } \right)\left( {\sqrt {50} - \sqrt {14} } \right)\]
Now, we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
$\Rightarrow $\[4ac = - 16\left( {\sqrt {50} + \sqrt {14} } \right)\left( {\sqrt {50} - \sqrt {14} } \right) = - 16\left( {{{\left( {\sqrt {50} } \right)}^2} - {{\left( {\sqrt {14} } \right)}^2}} \right)\]
So, \[4ac = - 16\left( {50 - 14} \right) = - 576\]
Now, roots of the equation:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
$\Rightarrow $ \[x = \dfrac{{ - 10 \pm \sqrt {100 - \left( { - 576} \right)} }}{{2\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
$\Rightarrow $\[x = \dfrac{{ - 10 \pm \sqrt {676} }}{{2\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
$\Rightarrow $ \[x = \dfrac{{ - 10 \pm 26}}{{2\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
$\Rightarrow $ \[x = \dfrac{{ - 36}}{{2\left( {\sqrt {200} + \sqrt {56} } \right)}},x = \dfrac{{16}}{{2\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
$\Rightarrow $\[x = \dfrac{{ - 18}}{{\left( {\sqrt {200} + \sqrt {56} } \right)}},x = \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
Now, the positive roots is \[x = \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}\]
Now, rationalizing the denominator of the root, as no option matches with this form,
$\Rightarrow $\[x = \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}} = \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}} \times \dfrac{{\left( {\sqrt {200} - \sqrt {56} } \right)}}{{\left( {\sqrt {200} - \sqrt {56} } \right)}}\]
$\Rightarrow $\[x = \dfrac{{8\left( {\sqrt {200} - \sqrt {56} } \right)}}{{\left( {\sqrt {200} + \sqrt {56} } \right)\left( {\sqrt {200} - \sqrt {56} } \right)}}\]
Now, we know that, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
$\Rightarrow $\[x = \dfrac{{8\left( {\sqrt {200} - \sqrt {56} } \right)}}{{\left( {200 - 56} \right)}}\]
$\Rightarrow $\[x = \dfrac{{8\left( {\sqrt {200} - \sqrt {56} } \right)}}{{144}} = \dfrac{{\left( {\sqrt {200} - \sqrt {56} } \right)}}{{18}}\]
$\Rightarrow $\[x = \dfrac{{\left( {10\sqrt 2 - 2\sqrt {14} } \right)}}{{18}}\]
$\Rightarrow $\[x = \dfrac{{5\sqrt 2 - \sqrt {14} }}{9}\]

Hence, the positive root of the equation is \[\dfrac{{5\sqrt 2 - \sqrt {14} }}{9}\], so option C is correct.

Note: Students must remember the quadratic formula. Also, after getting the root from the quadratic formula, they must rationalize the denominator or numerator part whichever is irrational to get the correct answer. If there is an option of none of the above, select it if you are sure none of the option is equal to the calculated root. Another approach towards solving the question can be by factorization method.
$\left( {\sqrt {200} + \sqrt {56} } \right){x^2} + 10x - 2\left( {\sqrt {50} - \sqrt {14} } \right) = 0$
$ $\Rightarrow $ \left( {\sqrt {200} + \sqrt {56} } \right){x^2} + 10x - \left( {2\sqrt {50} - 2\sqrt {14} } \right) = 0$
$ $\Rightarrow $ \left( {\sqrt {200} + \sqrt {56} } \right){x^2} + 10x - \left( {\sqrt {200} - \sqrt {56} } \right) = 0$
$ $\Rightarrow $ {x^2} + \dfrac{{10}}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}x - \dfrac{{\left( {\sqrt {200} - \sqrt {56} } \right)}}{{\left( {\sqrt {200} + \sqrt {56} } \right)}} = 0$
$ $\Rightarrow $ {x^2} + \dfrac{{18}}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}x - \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}x - \dfrac{{\left( {\sqrt {200} - \sqrt {56} } \right)}}{{\left( {\sqrt {200} + \sqrt {56} } \right)}} = 0$
$ $\Rightarrow $ x\left( {x + \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}} \right) - \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}\left( {x + \dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8}} \right) = 0$
Now,
$\dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8} = \dfrac{{\left( {\sqrt {200} - \sqrt {56} } \right)\left( {\sqrt {200} + \sqrt {56} } \right)}}{{8\left( {\sqrt {200} + \sqrt {56} } \right)}}$
Now, we know that, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
$ $\Rightarrow $ \dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8} = \dfrac{{{{\left( {\sqrt {200} } \right)}^2} - {{\left( {\sqrt {56} } \right)}^2}}}{{8\left( {\sqrt {200} + \sqrt {56} } \right)}}$
$ $\Rightarrow $ \dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8} = \dfrac{{200 - 56}}{{8\left( {\sqrt {200} + \sqrt {56} } \right)}}$
$ $\Rightarrow $ \dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8} = \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}$
So,
$x\left( {x + \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}} \right) - \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}\left( {x + \dfrac{{\sqrt {200} - \sqrt {56} }}{{\sqrt {200} + \sqrt {56} }} \times \dfrac{{\sqrt {200} + \sqrt {56} }}{8}} \right) = 0$
Becomes,
$x\left( {x + \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}} \right) - \dfrac{8}{{\left( {\sqrt {200} + \sqrt {56} } \right)}}\left( {x + \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}} \right) = 0$
\[ $\Rightarrow $ \left( {x - \dfrac{8}{{\sqrt {200} + \sqrt {56} }}} \right)\left( {x + \dfrac{{18}}{{\sqrt {200} + \sqrt {56} }}} \right) = 0\]
Thus, the roots of the above equations are,
\[x = \dfrac{8}{{\sqrt {200} + \sqrt {56} }},x = \dfrac{{ - 18}}{{\sqrt {200} + \sqrt {56} }}\]