(x+2)^2+(x-3)^2=53 solve
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(x+2)^2+(x-3)^2=53
x^2+4x+4+x^2-6x+9=53
2x^2-2x+13=53
2x^2-2x-40=0
2x^2-10x+8x-40=0
2x (x-5)+8 (x-5)=0
(x-5)(2x+8)=0
x-5=0. or. 2x+8=0
x=5. or. x=-4
x^2+4x+4+x^2-6x+9=53
2x^2-2x+13=53
2x^2-2x-40=0
2x^2-10x+8x-40=0
2x (x-5)+8 (x-5)=0
(x-5)(2x+8)=0
x-5=0. or. 2x+8=0
x=5. or. x=-4
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Answered by
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HEYA FRND. ..........☺
(x+2)^2 + (x-3)^2 = 53
x^2 + 4 + 4x + x^2 + 9 - 6x = 53
= 2x^2 - 2x + 13 = 53
So,
2x^2 - 2x - 40
2x^2 - 10x + 8x - 40
=> 2x (x- 5) + 8 (x- 5)
=> (x - 5) (2x + 8)
Therefore,
x = 5 , x = - 4 ..................ANS
____________________
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PLZ MARK MY ANSWER AS BRAINLIEST IF U LIKE IT. . . . . . . . . ☺
(x+2)^2 + (x-3)^2 = 53
x^2 + 4 + 4x + x^2 + 9 - 6x = 53
= 2x^2 - 2x + 13 = 53
So,
2x^2 - 2x - 40
2x^2 - 10x + 8x - 40
=> 2x (x- 5) + 8 (x- 5)
=> (x - 5) (2x + 8)
Therefore,
x = 5 , x = - 4 ..................ANS
____________________
HOPE IT WILL HELP YOU. . . . . . .
PLZ MARK MY ANSWER AS BRAINLIEST IF U LIKE IT. . . . . . . . . ☺
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