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

AcademicMathematicsNCERTClass 10

To do:

We have to find the roots of the given quadratic equations by applying the quadratic formula.

Solution:

(i) $2x^2-7x + 3 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 2, b = -7$ and $c =  3$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(-7)^{2}-4 \times 2\times(3)$

$=49-24$

$=25$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-7)+\sqrt{25}}{2(2)}$

$=\frac{7+5}{4}$

$=\frac{12}{4}$

$=3$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-7)-\sqrt{25}}{2(2)}$

$=\frac{7-5}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Hence, the roots of the given quadratic equation are $\frac{1}{2}, 3$.

(ii) $2x^2-x + 4 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 2, b = 1$ and $c = - 4$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(1)^{2}-4 \times 2(-4)$

$=1+32$

$=33$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-1+\sqrt{33}}{4}$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-1-\sqrt{33}}{4}$

Hence, the roots of the given quadratic equation are $\frac{-1+\sqrt{33}}{4}, \frac{-1-\sqrt{33}}{4}$.

(iii) $4x^2 - 4\sqrt3x + 3 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 4, b = -4\sqrt3$ and $c = 3$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(4 \sqrt{3})^{2}-4 \times 4 \times 3$

$=48-48$

$=0$

$\mathrm{D}=0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha=\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-4 \sqrt{3}+0}{8}$

$=\frac{-4 \sqrt{3}}{8}$

$=\frac{-\sqrt{3}}{2}$

$\beta=\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-4 \sqrt{3}-0}{8}$

$=\frac{-4 \sqrt{3}}{8}$

$=\frac{-\sqrt{3}}{2}$

Hence, the roots of the given quadratic equation are $\frac{-\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}$.

(iv) $2x^2 + x + 4 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 2, b = 1$ and $c = 4$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(1)^{2}-4 \times 2 \times 4$

$=1-32$

$=-31$

$\mathrm{D}<0$

Therefore, real roots do not exist.

raja
Updated on 10-Oct-2022 13:20:12

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