# Which of the following equations have two distinct real roots.

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A.2{x^2} - 3\sqrt 2 x + \dfrac{9}{4} = 0 \\

B.{x^2} + x - 5 = 0 \\

C.{x^2} + 3x + 2\sqrt 2 = 0 \\

D.5{x^2} - 3x + 1 = 0 \\

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**Hint:**In order to solve this problem we will apply the SriDharacharya formula in all the equations given and then we will observe the roots of the equations and will give answers accordingly. The formula of sridharacharya is$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Knowing this will solve your problem and will give you the right answers.

__Complete step-by-step answer__:We know that the two roots of the equation $a{x^2} + bx + c = 0$ can be found from the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

So, the first equation is $2{x^2} - 3\sqrt 2 x + \dfrac{9}{4} = 0$

So, the roots of the equation according to the formula above is $x = \dfrac{{3\sqrt 2 \pm \sqrt {{{\left( { - 3\sqrt 2 } \right)}^2} - 4\left( 2 \right)\left( {\dfrac{9}{4}} \right)} }}{{2\left( 2 \right)}} = \dfrac{{3\sqrt 2 }}{4}$

Hence both roots of this equation are the same and that is $\dfrac{{3\sqrt 2 }}{4}$.

The second equation is ${x^2} + x - 5 = 0$

So, the roots of the equation according to the formula above is $x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 5} \right)} }}{{2\left( 1 \right)}} = \dfrac{{ - 1 \pm \sqrt {21} }}{2}$

Hence the two roots of this equation are is $\dfrac{{ - 1 + \sqrt {21} }}{2}$ and $\dfrac{{ - 1 - \sqrt {21} }}{2}$.

Here you can clearly see that the roots are different but not real as they have negative value under root.

The third equation is ${x^2} + 3x + 2\sqrt 2 = 0$

So, the roots of the equation according to the formula above is $x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 1 \right)\left( {2\sqrt 2 } \right)} }}{{2\left( 1 \right)}} = \dfrac{{ - 3 \pm \sqrt {9 - 8\sqrt 2 } }}{2}$

Hence the two roots of this equation is different and those are is $\dfrac{{ - 3 + \sqrt {9 - 8\sqrt 2 } }}{2}$ and $\dfrac{{ - 3 - \sqrt {9 - 8\sqrt 2 } }}{2}$.

The fourth equation is $5{x^2} - 3x + 1 = 0$

So, the roots of the equation according to the formula above is $x = \dfrac{{3 \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 5 \right)\left( 1 \right)} }}{{2\left( 5 \right)}} = \dfrac{{3 \pm \sqrt {9 - 20} }}{{10}} = \dfrac{{3 \pm \sqrt { - 11} }}{{10}} = \dfrac{{3 \pm 11i}}{{10}}$

Hence the two roots of this equation is different and those are is $\dfrac{{3 + 11i}}{{10}}$ and $\dfrac{{3 - 11i}}{{10}}$.

So, we can clearly see that the equations of option B and C have different and real roots. While A has the same roots but real and D have different roots but imaginary.

**So, the answer is B.**

**Note**:When you get to solve such problems you need to know that roots of quadratic equation can be found by the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and it can give you real and imaginary roots may be distinct or same according to the equation present. The discriminant is the part of the quadratic formula underneath the square root symbol: $\sqrt {{b^2} - 4ac} $. The discriminant tells us whether there are two solutions, one solution, or no solutions. With the help of discriminant we can find out whether the quadratic equations are real or imaginary. Knowing this will help you further and solve your problem.