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Answers
Step-by-step explanation:
Solution :-
1)
Given equation is x²-3x = 4
=> x²-3x -4 = 0
=> x²+x-4x-4 = 0
=> x(x+1)-4(x+1) = 0
=> (x+1)(x-4) = 0
=> x+1 = 0 or x-4 = 0
=> x = -1 or x = 4
2)
Given equation is x²+4x+4 = 0
=> x²+2x+2x+4 = 0
=> x(x+2)+2(x+2) = 0
=> (x+2)(x+2) = 0
=> x+2 = 0 or x+2 = 0
=> x = -2 or x = -2
3)
Given equation is 2x²+x-28 = 0
=> 2x²+8x-7x-28 = 0
=> 2x(x+4)-7(x+4) = 0
=> (x+4)(2x-7) = 0
=> x+4 = 0 or 2x-7 = 0
=> x = -4 or 2x = 7
=> x = -4 or x = 7/2
4)
Given equation is x²-8x = -16
=> x²-8x+16 = 0
=> x²-4x-4x+16 = 0
=> x(x-4)-4(x-4) = 0
=> (x-4)(x-4) = 0
=> x-4 = 0 or x-4 = 0
=> x = 4 or x = 4
5)
Given equation is x²+2x+1 = 0
=> x²+x+x+1 = 0
=> x(x+1)+1(x+1) = 0
=> (x+1)(x+1) = 0
=> x+1 = 0 or x+1 = 0
=> x = -1 or x = -1
6)
Given equation is 4x²-24x-36 = 0
=> 4(x²-6x-9) = 0
=> x²-6x-9 = 0
=> x²-2(x)(3) = 9
On adding 3² both sides then
=> x²-2(3)x)+3² = 9+9
=> (x-3)² = 18
=> (x-3) = ±√18
=> x-3 = ±√(2×9)
=> x-3 = ±3√2
=> x = 3±3√2
=> x = 3+3√2 or 3-3√2
1-P
2-F
3-I
4-E
5-A
6-R
a) I Found the solution for the each of equations by factorization method and completing the square method.
b)
→ I uses the basic concepts for splitting the middle terms
→ I used the factorization method
→ I used completing the square method
→ I applied this equation by transposing the terms from LHS and RHS .