Answer:
See explanation
Step-by-step explanation:
The place to begin in attempting to solve a quadratic equation is to try to factorize it. If it can not be factorized, then look at the side of equation that contains the variable (x,b,m,n and so on). If that side is a perfect square, try to take square root of both sides.
Also, if the coefficient of the squared term in the equation is 1 and the coefficient of the middle term bearing the variable is an even number, then you can attempt to use completing the square method. Formula method can be adopted as a last resort.
x^2 -4x + 4 =0 can be solved by factorization
X(x-2) - 2(x-2) =0
Hence;
(x-2) (x-2) =0
x= 2 (twice)
(2x-2)^2 = 4 can be solved by taking square root
√[(2x-2)^2] = √4
2x-2 = 2
x= ±2
x^2 + 6x - 11 = 0 can be solved by completing the square
(x + 3)^2 = 11 + 3^2
(x + 3)^2 = 20 taking square root of both sides
x + 3 = ±√20
x + 3 = ±2√5
x= -3 ±2√5
2x^2 - x + 5 = 0 can be solved by formula method
Using
x= -b±√b^2 - 4ac/2a
a= 2, b = -1, c= 5
x= 1 ± √1 - 40/4
x= 1 ± √-39/4
x= 1 ± 39i/4
Answer:
x2+y2−4x+20y+4=0.
Explanation:
The eqn. of a circle with centre (h,k) and radius r is given by,
(x−h)2+(y−k)2=r2.
Accordingly, the reqd. eqn. is,
(x−2)2+(y+10)2=3^2, i.e.,
x2+y2−4x+20y+4=0.
Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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