Question

Read the instructions first and then answer the 1 and 2 follow the instructions.

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Answer 1

Answer:

which formula is I'm using in the box

highlighted

(I) ( a - b )²

(ii) a² - b²

Answer 2

$\large\underline{\sf{Solution-1}}$

The given quadratic equation is

$\red{\rm :\longmapsto\: {2x}^{2} - 10x + 8 = 0}$

Ayesha use the following steps for completing squares

$\red{\rm :\longmapsto\: {2x}^{2} - 10x = - 8}$

$\red{\rm :\longmapsto\: {2x}^{2} - 10x + 25 = - 8 + 25}$

Ayesha now added and subtract 25 to make it complete squares which is wrong.

As to add and subtract the square of half the coefficient of x, we have to first keep the coefficient of x^2 unity.

And Ayesha can't make the coefficient of x^2 unity.

So, she can't arrived to proper solution.

The correct steps are as follow :-

$\red{\rm :\longmapsto\: {2x}^{2} - 10x + 8 = 0}$

$\red{\rm :\longmapsto\: {x}^{2} - 5x + 4 = 0}$

$\red{\rm :\longmapsto\: {x}^{2} - 5x = - 4}$

$\red{\rm :\longmapsto\: {x}^{2} - 5x + \dfrac{25}{4} = - 4 + \dfrac{25}{4} }$

$\red{\rm :\longmapsto\: {\bigg[x - \dfrac{5}{2} \bigg]}^{2} = \dfrac{ - 16 + 25}{4} \: }$

$\red{\rm :\longmapsto\: {\bigg[x - \dfrac{5}{2} \bigg]}^{2} = \dfrac{9}{4} \: }$

$\red{\rm :\longmapsto\: {\bigg[x - \dfrac{5}{2} \bigg]}^{2} = {\bigg[\dfrac{3}{2} \bigg]}^{2} \: }$

$\rm :\longmapsto\:x - \dfrac{5}{2} \: = \: \pm \: \dfrac{3}{2}$

$\rm :\longmapsto\:x = \dfrac{5}{2} \: \pm \: \dfrac{3}{2}$

$\red{\rm \implies\:x = 4 \: \: or \: \: x = 1}$

$\green{\large\underline{\sf{Solution-2}}}$

The given quadratic equation is

$\rm :\longmapsto\: {9x}^{2} - 36 = 0$

Chris want to use method of Completing squares to solve this quadratic equation.

But he cannot use the Method of Completing Squares to solve this quadratic equation problem as coefficient of x is 0.

So, addition and Subtracting of 0, make it nothing.

Rather, he can use factorization method to solve this problem.

$\rm :\longmapsto\: 9({x}^{2} - 4)= 0$

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

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

$\rm :\longmapsto\: (x + 2)(x - 2) = 0$

$\green{\rm \implies\:x = 2 \: \: or \: \: x = - \: 2}$

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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