Q2.    Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answers (1)
D Divya Prakash Singh

(i) 2x^2-7x+3 = 0

The general form of a quadratic equation is : ax^2+bx+c = 0, where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

a = 2,\ b = -7\ c = 3

And the quadratic formula for finding the roots is:

x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{7 \pm \sqrt{49-24}}{4}

\Rightarrow x= \frac{7 \pm 5}{4}

\Rightarrow x= \frac{7 + 5}{4} = 3\ or\ x= \frac{7 - 5}{4} = \frac{1}{2}

Therefore, the real roots are: x =3,\ \frac{1}{2}

 

(ii) 2x^2+x-4 = 0

The general form of a quadratic equation is : ax^2+bx+c = 0, where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

a = 2,\ b = 1\ c =-4

And the quadratic formula for finding the roots is:

x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-1 \pm \sqrt{1+32}}{4}

\Rightarrow x= \frac{-1 \pm \sqrt{33}}{4}

\Rightarrow x= \frac{-1 + \sqrt{33}}{4} \ or\ x= \frac{-1 - \sqrt{33}}{4}

Therefore, the real roots are: x = \frac{-1+\sqrt{33}}{4}\ or\ \frac{-1-\sqrt{33}}{4}

 

 

(iii) 4x^2+4\sqrt3x+3 = 0

The general form of a quadratic equation is : ax^2+bx+c = 0, where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

a = 4,\ b = 4\sqrt{3}\ c =3

And the quadratic formula for finding the roots is:

x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}

\Rightarrow x= \frac{-4\sqrt{3} \pm 0}{8}

Therefore, the real roots are: x = \frac{-\sqrt{3}}{2}\ or\ \frac{-\sqrt{3}}{2}

 

 

(iv) 2x^2+x+4 = 0

The general form of a quadratic equation is : ax^2+bx+c = 0, where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

a = 2,\ b = 1,\ c =4

And the quadratic formula for finding the roots is:

x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-1 \pm \sqrt{1-32}}{4}

\Rightarrow x= \frac{-1 \pm \sqrt{-31}}{4}

 Here the term inside the root is negative

Therefore there are no real roots for the given equation.

 

 

 

 

 

 

 

Exams
Articles
Questions