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

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

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(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.

 

 

 

 

 

 

 

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