Solve: 13-x²=(x+5)²
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13-x^2=x^2+35+10x
2x^2+10x+22
d=b^2-4ac
100-169
there as d is less than zero this equation doesnot have real roota
2x^2+10x+22
d=b^2-4ac
100-169
there as d is less than zero this equation doesnot have real roota
chanpreet300:
sorry wait
it is 25 not 35
2X^2+10x+12
2x^2+6x+4x+12
2x(x+3)+4(x+3)
(x+3)(2x+4)
x=-3 x=-2
2x^2+6x+4x+12
2x(x+3)+4(x+3)
(x+3)(2x+4)
x=-3 x=-2
Answered by
6
Heya,
=> 13 - x² = (x + 5)²
=> 13 - x² = x² + 25 + 10x
=> x² + x² + 25 - 13 + 10x = 0
=> 2x² + 12 + 10x = 0
=> 2(x² + 5x + 6) = 0
=> x² + 5x + 6 = 0
=> x² + 2x + 3x + 6 = 0
=> x(x + 2) + 3(x + 2) = 0
=> (x + 2)(x + 3) = 0
=> x = -2, -3. ..... Answer
Hope this helps.....:)
=> 13 - x² = (x + 5)²
=> 13 - x² = x² + 25 + 10x
=> x² + x² + 25 - 13 + 10x = 0
=> 2x² + 12 + 10x = 0
=> 2(x² + 5x + 6) = 0
=> x² + 5x + 6 = 0
=> x² + 2x + 3x + 6 = 0
=> x(x + 2) + 3(x + 2) = 0
=> (x + 2)(x + 3) = 0
=> x = -2, -3. ..... Answer
Hope this helps.....:)
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