Question

★ Quadratic Equation ★

1: Find the roots of the following quadratic equations by the method of completing the squares :

• $\boxed{\sf{\red{ 4x² \:+\: 4 \sqrt{3}x \:+\: 3 \:=\: 0 }}}$

• $\boxed{\sf{\red{ 2x² \:+\: x \:-\: 4 \:=\: 0 }}}$
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer :- 1

Given Quadratic Equation is

$\rm :\longmapsto\: {4x}^{2} + 4 \sqrt{3}x + 3 = 0$

Step :- 1 Shift the constant term on RHS.

$\rm :\longmapsto\: {4x}^{2} + 4 \sqrt{3}x= - \: 3$

Step :- 2 Make the coefficient of x² unity, So Divide both sides by 4.

$\rm :\longmapsto\: {x}^{2} +\sqrt{3}x= - \: \dfrac{3}{4}$

Step :- 3 Add square of half the coefficient of x on both sides,

$\rm :\longmapsto\: {x}^{2} +\sqrt{3}x + {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{2} = - \: \dfrac{3}{4} + {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{2}$

$\rm :\longmapsto\: {x}^{2} +2 \times\dfrac{ \sqrt{3} }{2} \times x + {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{2} = - \:\dfrac{3}{4}+\dfrac{3}{4}$

$\rm :\longmapsto\:{\bigg(x + \dfrac{ \sqrt{3} }{2} \bigg) }^{2} = 0$

$\rm :\longmapsto\:{\bigg(x + \dfrac{ \sqrt{3} }{2} \bigg) } = 0$

$\bf\implies \:x = - \: \dfrac{ \sqrt{3} }{2}$

Answer :- 2

Given Quadratic Equation is

$\rm :\longmapsto\: {2x}^{2} + x - 4 = 0$

Step :- 1 Shift the constant term on RHS.

$\rm :\longmapsto\: {2x}^{2} + x = 4$

Step :- 2 Make the coefficient of x² unity, So Divide both sides by 2.

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{2} x = 2$

Step :- 3 Add square of half the coefficient of x on both sides,

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{2} x + {\bigg(\dfrac{1}{4} \bigg) }^{2} = 2 + {\bigg(\dfrac{1}{4} \bigg) }^{2}$

$\rm :\longmapsto\: {x}^{2} +2 \times \dfrac{1}{4} x + {\bigg(\dfrac{1}{4} \bigg) }^{2} = 2 + {\bigg(\dfrac{1}{16} \bigg) }$

$\rm :\longmapsto\:{\bigg(x + \dfrac{1}{4}\bigg) }^{2} = \dfrac{32 + 1}{16}$

$\rm :\longmapsto\:{\bigg(x + \dfrac{1}{4}\bigg) }^{2} = \dfrac{33}{16}$

$\rm :\longmapsto\:{\bigg(x + \dfrac{1}{4}\bigg) }= \pm \: \dfrac{ \sqrt{33} }{4}$

$\rm :\longmapsto\:x= - \: \dfrac{1}{4} \pm \: \dfrac{ \sqrt{33} }{4}$

$\bf\implies \:x = \dfrac{ - 1 + \sqrt{33} }{4} \: \: \: or \: \: \: \dfrac{ - 1 - \sqrt{33} }{4}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac