Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) $2x^2 - 7x + 3 = 0$
(ii) $2x^2 + x - 4 = 0$
(iii) $4x^2 + 4\sqrt3x + 3 = 0$
(iv) $2x^2 + x + 4 = 0$

AcademicMathematicsNCERTClass 10

To do:

We have to find the roots of the given quadratic equations by the method of completing the square.

Solution:

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

$2(x^2 - \frac{7}{2} x +\frac{3}{2}) = 0$  

$x^2 - \frac{7}{2} x +\frac{3}{2} = 0$  

$x^2 - 2\times \frac{1}{2}\times \frac{7}{2} x = -\frac{3}{2}$  

$x^2 - 2\times \frac{7}{4} x = -\frac{3}{2}$  

Adding $(\frac{7}{4})^2$ on both sides completes the square. Therefore,

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

$(x-\frac{7}{4})^2=-\frac{3}{2}+\frac{49}{16}$      (Since $(a-b)^2=a^2-2ab+b^2$)

$(x-\frac{7}{4})^2=\frac{49-3\times8}{16}$

$(x-\frac{7}{4})^2=\frac{49-24}{16}$

$(x-\frac{7}{4})^2=\frac{25}{16}$

$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}$     (Taking square root on both sides)

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

$x=\frac{7}{4}+\frac{5}{4}$ or $x=\frac{7}{4}-\frac{5}{4}$

$x=\frac{7+5}{4}$ or $x=\frac{7-5}{4}$

$x=\frac{12}{4}$ or $x=\frac{2}{4}$

$x=3$ or $x=\frac{1}{2}$

The values of $x$ are $3$ and $\frac{1}{2}$.

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

$2(x^2 + \frac{1}{2} x -\frac{4}{2}) = 0$  

$x^2 + \frac{1}{2} x -2 = 0$  

$x^2 + 2\times \frac{1}{2}\times \frac{1}{2} x = 2$  

$x^2 + 2\times \frac{1}{4} x = 2$  

Adding $(\frac{1}{4})^2$ on both sides completes the square. Therefore,

$x^2 + 2\times (\frac{1}{4}) x + (\frac{1}{4})^2 = 2+(\frac{1}{4})^2$

$(x+\frac{1}{4})^2=2+\frac{1}{16}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{1}{4})^2=\frac{1+2\times16}{16}$

$(x+\frac{1}{4})^2=\frac{1+32}{16}$

$(x+\frac{1}{4})^2=\frac{33}{16}$

$x+\frac{1}{4}=\pm \sqrt{\frac{33}{16}}$     (Taking square root on both sides)

$x=\sqrt{\frac{33}{16}}-\frac{1}{4}$ or $x=-\sqrt{\frac{33}{16}}-\frac{1}{4}$

$x=\frac{\sqrt{33}-1}{4}$ or $x=-(\frac{\sqrt{33}+1}{4})$

The values of $x$ are $\frac{\sqrt{33}-1}{4}$ and $-(\frac{\sqrt{33}+1}{4})$.

(iii) $4x^2 + 4\sqrt3 x + 3 = 0$

$4(x^2 + \sqrt3 x +\frac{3}{4})=0$  

$x^2+2\times \frac{1}{2} \times \sqrt3 x =-\frac{3}{4}$

$x^2+2\frac{\sqrt3}{2}x=-\frac{3}{4}$

Adding $(\frac{\sqrt3}{2})^2$ on both sides completes the square. Therefore,

$x^2+2\frac{\sqrt3}{2}x+(\frac{\sqrt3}{2})^2=-\frac{3}{4}+(\frac{\sqrt3}{2})^2$

$(x+\frac{\sqrt3}{2})^2=-\frac{3}{4}+\frac{3}{4}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{\sqrt3}{2})^2=0$

$x+\frac{\sqrt3}{2}=0$

$x=-\frac{\sqrt3}{2}$ or $x=-\frac{\sqrt3}{2}$

The values of $x$ are $-\frac{\sqrt3}{2}$ and $-\frac{\sqrt3}{2}$.

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

$2(x^2 + \frac{1}{2} x +\frac{4}{2}) = 0$  

$x^2 + \frac{1}{2} x + 2 = 0$  

$x^2 + 2\times \frac{1}{2}\times \frac{1}{2} x = -2$  

$x^2 + 2\times \frac{1}{4} x = -2$  

Adding $(\frac{1}{4})^2$ on both sides completes the square. Therefore,

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

$(x+\frac{1}{4})^2=-2+\frac{1}{16}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{1}{4})^2=\frac{1-2\times16}{16}$

$(x+\frac{1}{4})^2=\frac{1-32}{16}$

$(x+\frac{1}{4})^2=\frac{-31}{16}$

$x+\frac{1}{4}=\pm \sqrt{\frac{-31}{16}}$     (Taking square root on both sides)

$x=\sqrt{\frac{-31}{16}}-\frac{1}{4}$ or $x=-\sqrt{\frac{-31}{16}}-\frac{1}{4}$

$x=\frac{\sqrt{-31}-1}{4}$ or $x=-(\frac{\sqrt{-31}+1}{4})$

Therefore, no real roots exist for the given quadratic equation. 

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

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