Please solve q no 1 full

1.	Please solve q no 1 full
Answer» Concept : Quadratic equations are of the form : ⠀⠀⠀⌬ ax² + bx + c = 0 where , a ≠ 0 and a, b, c are real numbers. So now for the given questions we would simplify each one of them and see whether it can be simplified in the general form of quadratic equation. ‎ ‎ QUESTION (i) : ⠀⠀⠀⌬ $\sf\:x^{2}-2x=(-2) (3-x)$ ‎ Solution : $\sf\:x^{2}-2x=(-2) (3-x)$ $\sf\implies\:x^{2}-2x=(-6+2x)$ $\sf\implies\:x^{2}-2x=-6+2x$ $\sf\implies\:x^{2}-2x-2x=-6$ $\sf\implies\:x^{2}-4x=-6$ $\sf\implies\:x^{2}-4x+-6=0$ ‎ $\sf\therefore\:It~is~a~quadratic~equation$ ════════════════════════ Question (II) : ⠀⠀⠀⌬ $\sf\:(x-3) (2x+1) =x(x+5)$ ‎ Solution : $\sf\:(x-3) (2x+1) =x(x+5)$ $\sf\implies\:x(2x+1)-3(2x+1) =x(x+5)$ $\sf\implies\:2x^{2}+x-6x-3=x^{2}+5x$ $\sf\implies\:2x^{2}-5x-3=x^{2}+5x$ $\sf\implies\:2x^{2}-5x+5x-3=x^{2}$ $\sf\implies\:2x^{2}-\cancel{5x}+\cancel{5x}-3=x^{2}$ $\sf\implies\:2x^{2}-3=x^{2}$ $\sf\implies\:2x^{2}-x^{2}-3=0$ $\sf\implies\:x^{2}-3=0$ ‎ $\sf\therefore\:It~is~a~quadratic~equation$ ════════════════════════ Question (iii) : ‎ ⠀⠀⠀⌬ $\sf\:(2x-1) (x-3) =(x+5) (x-1)$ ‎ Solution : $\sf\:(2x-1) (x-3) =(x+5) (x-1)$ $\sf\implies\:2x (x-3) -1(x-3) =x(x-1)+5(x-1)$ $\sf\implies\:2x^{2}-6x-x+3=x^{2}-x+5x-5$ $\sf\implies\:2x^{2}-7x+3=x^{2}+4x-5$ $\sf\implies\:2x^{2}-x^{2}-7x-4x+3+5=0$ $\sf\implies\:x^{2}-11x+8=0$ ‎ $\sf\therefore\:It~is~a~quadratic~equation$ ════════════════════════ Question (iv) : ⠀⠀⠀⌬ $\sf\:x^{2}+3x+1=(x-2) ^{2}$ ‎ Solution : $\sf\:x^{2}+3x+1=(x-2) ^{2}$ $\sf\implies\:x^{2}+3x+1=x^{2}-2 \times x \times 2 +(2)^{2}$ $\sf\implies\:x^{2}+3x+1=x^{2}-4x +4$ $\sf\implies\:x^{2}-x^{2}+3x+4x+1-4=0$ $\sf\implies\:\cancel{x^{2}}-\cancel{x^{2}}+3x+4x+1-4=0$ $\sf\implies\:3x+4x+1-4=0$ $\sf\implies\:7x+1-4=0$ $\sf\implies\:7x-3=0$ ‎ $\sf\therefore\:It~is~not~a~quadratic~equation$ [ Since a = 0 ] ════════════════════════ Question (V) : ‎ ⠀⠀⠀⌬ $\sf\:(x+2) ^{3}=2x(x^{2}-1)$ ‎ Solution : $\sf\:(x+2) ^{3}=2x(x^{2}-1)$ $\sf\implies\:x^{3}+(2)^{3}+(3 \times x \times 2)(x+2)=2x^{3}-2x$ $\sf\implies\:x^{3}+8+6x(x+2)=2x^{3}-2x$ $\sf\implies\:x^{3}+8+6x^{2}+12x=2x^{3}-2x$ $\sf\implies\:x^{3}-2x^{3}+8+6x^{2}+12x+2x=0$ $\sf\implies\:-x^{3}+8+6x^{2}+14x=0$ $\sf\implies\:-x^{3}+6x^{2}+14x+8=0$ ‎ $\sf\therefore\:It~is~not~a~quadratic~equation$ [ Since , highest DEGREE of the equation is 3 ] ════════════════════════ ‎Note : Scroll left to right to VIEW the answer properly !~ ‎

Answer»

Concept :

Quadratic equations are of the form :

⠀⠀⠀⌬ ax² + bx + c = 0

where , a ≠ 0 and a, b, c are real numbers.

So now for the given questions we would simplify each one of them and see whether it can be simplified in the general form of quadratic equation.

‎

QUESTION (i) :

⠀⠀⠀⌬ $\sf\:x^{2}-2x=(-2) (3-x)$

‎

Solution :

$\sf\:x^{2}-2x=(-2) (3-x)$

$\sf\implies\:x^{2}-2x=(-6+2x)$

$\sf\implies\:x^{2}-2x=-6+2x$

$\sf\implies\:x^{2}-2x-2x=-6$

$\sf\implies\:x^{2}-4x=-6$

$\sf\implies\:x^{2}-4x+-6=0$

‎

$\sf\therefore\:It~is~a~quadratic~equation$

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Question (II) :

⠀⠀⠀⌬ $\sf\:(x-3) (2x+1) =x(x+5)$

‎

Solution :

$\sf\:(x-3) (2x+1) =x(x+5)$

$\sf\implies\:x(2x+1)-3(2x+1) =x(x+5)$

$\sf\implies\:2x^{2}+x-6x-3=x^{2}+5x$

$\sf\implies\:2x^{2}-5x-3=x^{2}+5x$

$\sf\implies\:2x^{2}-5x+5x-3=x^{2}$

$\sf\implies\:2x^{2}-\cancel{5x}+\cancel{5x}-3=x^{2}$

$\sf\implies\:2x^{2}-3=x^{2}$

$\sf\implies\:2x^{2}-x^{2}-3=0$

$\sf\implies\:x^{2}-3=0$

‎

$\sf\therefore\:It~is~a~quadratic~equation$

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Question (iii) :

‎

⠀⠀⠀⌬ $\sf\:(2x-1) (x-3) =(x+5) (x-1)$

‎

Solution :

$\sf\:(2x-1) (x-3) =(x+5) (x-1)$

$\sf\implies\:2x (x-3) -1(x-3) =x(x-1)+5(x-1)$

$\sf\implies\:2x^{2}-6x-x+3=x^{2}-x+5x-5$

$\sf\implies\:2x^{2}-7x+3=x^{2}+4x-5$

$\sf\implies\:2x^{2}-x^{2}-7x-4x+3+5=0$

$\sf\implies\:x^{2}-11x+8=0$

‎

$\sf\therefore\:It~is~a~quadratic~equation$

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Question (iv) :

⠀⠀⠀⌬ $\sf\:x^{2}+3x+1=(x-2) ^{2}$

‎

Solution :

$\sf\:x^{2}+3x+1=(x-2) ^{2}$

$\sf\implies\:x^{2}+3x+1=x^{2}-2 \times x \times 2 +(2)^{2}$

$\sf\implies\:x^{2}+3x+1=x^{2}-4x +4$

$\sf\implies\:x^{2}-x^{2}+3x+4x+1-4=0$

$\sf\implies\:\cancel{x^{2}}-\cancel{x^{2}}+3x+4x+1-4=0$

$\sf\implies\:3x+4x+1-4=0$

$\sf\implies\:7x+1-4=0$

$\sf\implies\:7x-3=0$

‎

$\sf\therefore\:It~is~not~a~quadratic~equation$

[ Since a = 0 ]

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Question (V) :

‎

⠀⠀⠀⌬ $\sf\:(x+2) ^{3}=2x(x^{2}-1)$

‎

Solution :

$\sf\:(x+2) ^{3}=2x(x^{2}-1)$

$\sf\implies\:x^{3}+(2)^{3}+(3 \times x \times 2)(x+2)=2x^{3}-2x$

$\sf\implies\:x^{3}+8+6x(x+2)=2x^{3}-2x$

$\sf\implies\:x^{3}+8+6x^{2}+12x=2x^{3}-2x$

$\sf\implies\:x^{3}-2x^{3}+8+6x^{2}+12x+2x=0$

$\sf\implies\:-x^{3}+8+6x^{2}+14x=0$

$\sf\implies\:-x^{3}+6x^{2}+14x+8=0$

‎

$\sf\therefore\:It~is~not~a~quadratic~equation$

[ Since , highest DEGREE of the equation is 3 ]

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‎Note : Scroll left to right to VIEW the answer properly !~

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