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Sagot :
Solving Quadratic Equation by Completing the Square:
ALWAYS REMEMBER the standard form of a Quadratic Equation is Ax2 + Bx + C = 0. 1. 5p2 + 2p – 9 = 0
First ALWAYS make your A=1. In this case, you need to DIVIDE the whole equation by 5.
So, we can have:
(5p2 + 2p – 9)/5 = 0/5
p2 + 2/5 p – 9/5 = 0
p2 + 2/5 p = 9/5
Next get (B/2)2 and ADD it to the terms on the LEFT and RIGHT side of the equation.
(B/2)2 = {(2/5)(1/2)}2
= (1/5)2
(B/2)2 = 1/25
p2 + 2/5 p + 1/25 = 9/5 + 1/25
This time we have completed the square (the Left side of the equation is a perfect square; it can be factored completely) And we will have:
(p + 1/5)2 = (45+1)/25
(p + 1/5)2 = 46/25
Then take the square root of both sides of the equation for us to get the roots. We have:
p +1/ 5 = √(46/25)
p + 1/5 = (√46)/5
p = +/- (√46)/5 - 1/5
So you now have the roots that will satisfy your equation:
P1 = (√46)/5 - 1/5 p2 = - (√46)/5 - 1/5
P1 = {(√46)-1} / 5 ANSWER p2 = - {(√46)-1} / 5 ANSWER
ALWAYS REMEMBER the standard form of a Quadratic Equation is Ax2 + Bx + C = 0. 1. 5p2 + 2p – 9 = 0
First ALWAYS make your A=1. In this case, you need to DIVIDE the whole equation by 5.
So, we can have:
(5p2 + 2p – 9)/5 = 0/5
p2 + 2/5 p – 9/5 = 0
p2 + 2/5 p = 9/5
Next get (B/2)2 and ADD it to the terms on the LEFT and RIGHT side of the equation.
(B/2)2 = {(2/5)(1/2)}2
= (1/5)2
(B/2)2 = 1/25
p2 + 2/5 p + 1/25 = 9/5 + 1/25
This time we have completed the square (the Left side of the equation is a perfect square; it can be factored completely) And we will have:
(p + 1/5)2 = (45+1)/25
(p + 1/5)2 = 46/25
Then take the square root of both sides of the equation for us to get the roots. We have:
p +1/ 5 = √(46/25)
p + 1/5 = (√46)/5
p = +/- (√46)/5 - 1/5
So you now have the roots that will satisfy your equation:
P1 = (√46)/5 - 1/5 p2 = - (√46)/5 - 1/5
P1 = {(√46)-1} / 5 ANSWER p2 = - {(√46)-1} / 5 ANSWER
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