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Answers
Step-by-step explanation:
Solution :-
1. Given equation is 4x²-11x = 7
=> 4x²-11x-7 = 0
The standard form = 4x²-11x-7 = 0
It is in the form of ax²+bx+c = 0
We have,
a = 4
b = -11
c = -7
2.Given equation is 5x² = 3x
=> 5x²-3x = 0
The standard form = 5x²-3x +0 = 0 or 5x²-3x = 0
It is in the form of ax²+bx+c = 0
We have,
a = 5
b = -3
c = 0
3.Given equation is (x+3)(x-2) = 1
=> x(x-2) +3(x-2) = 1
=> x²-2x+3x-6 = 1
=> x²+x-6 = 1
=> x²+x-6-1 = 0
=> x²+x-7 = 0
The standard form = x²+x-7 = 0
It is in the form of ax²+bx+c = 0
We have,
a = 1
b = 1
c = -7
4. Given equation is (x-7)²+3 = 0
=> x²-14x+49+3 = 0
=> x²-14x+52 = 0
The standard form = x²-14x+52 = 0
It is in the form of ax²+bx+c = 0
We have,
a = 1
b = -14
c = 52
5.Given equation is 6x² = 5x-4
=> 6x²-5x+4 = 0
The standard form = 6x²-5x+4 = 0
It is in the form of ax²+bx+c = 0
We have,
a = 6
b = -5
c = 4
a)
→ Transposing the terms form LHS to RHS
→ (a-b)² = a²-2ab+b²
→ When transposing the positive sign one side of other side in the equation then it becomes negative quantity.
→ When transposing the negative sign one side of other side in the equation then it becomes positive quantity.
→ Multiplication of terms in the equation.
b)
→ There is no difficulty to write the standard form of the equation .
→ It is easy to write in the form of ax²+bx+c = 0.