Solve by completing the square.X^2-18x-4=0?
ok, add 4 on both sides x^2 -18x=4 so the formula to find the square is (b/2)^2 so add 81 on both sides x^2 -18x +81= 85 (x+9)^2=85 so find the square root of both sides x+9=+ sqrt 85 and - sqrt 85 so x=9+ sqrt 85 and 9 - sqrt 85
How do i factor by completing the square?
x⁴ − 10x²y⁴ + 9y⁸ = (x²)² − 6x²y⁴ − 4x²y⁴ + (3y⁴)² = (x²)² − 2(x²)(3y⁴) + (3y⁴)² − 4x²y⁴ = (x² − 3y⁴)² − 4x²y⁴ = (x² − 3y⁴)² − (2xy��)² = (x² − 3y⁴ + 2xy²) (x² − 3y⁴ − 2xy²) = (x² + 2xy² − 3y⁴) (x² − 2xy² − 3y⁴) = (x−y²)(x+3y²)(x+y²)(x−3y²) = (x−y²)(x+y²)(x−3y²)(x+3y²).
How do I factor x^2-38x+c by completing the square?
ok dude, they are asking you to complete the square for x2 - 38x by the way, i hate typing the ^ thing, so the 2 AFTER the letter means "squared" anyway, when you square the binomial (a - b)2, you get a2 - 2ab + b2 in other words, the number in the middle is TWICE as much as the numbers in front of the a and the b multiplied together. there is a "1" understood in front of the x2 so the original number in front of the x must also have been a "1" because the square root of 1 is one. if there had been a 9 in front of the x2, then the original number in front of the x would have been square root 9, or 3... ok? the 38 is TWICE the product of the original a and b numbers, so the product of the a and b numbers is 19, or half of 38. since the x number is 1, the "a" number is 1 and the "b" number has to be 19, because 1 x 19 = 19. so the original binomial was (x - 19)2 another example, lets say you had to complete the square of 9x2 - 24x for the reason explained above, the original x number is square root 9, or 3 for the reason explained above, the product of the 2 numbers is half 24, or 12 so, 3 times what = 12? the answer is 3 x 4 so the original binomial was (3x - 4)2 hope you get it now :-)))
3x² - 12x + 11 = 0 ; solve by completing the square?
Move the constant term (number by itself) to the right side. 3x^2 - 12x = -11 Factor out the first-term coefficient from the first two terms. 3(x^2 - 4x) = -11 Complete the trinomial square; for the last term of it divide the second-term coefficient by 2 (4/2 = 2) and square it (2^2 = 4). This times the number outside of the parentheses is being added to the left side of the equation; add it to the right side, also. 3(x^2 - 4x + 4) = -11 + 3*4 3(x - 2)^2 = -11 + 12 3(x - 2)^2 = 1 (x - 2)^2 = 1/3 x - 2 = +/- sqrt (1/3) x = 2 +/- sqrt (1/3) sqrt (1/3) = sqrt 1 / sqrt 3; multiply numerator and denominator each by sqrt 3 to rationalize the denominator ,,,,,,,,,,,,,,= (sqrt 1 * sqrt 3) / (sqrt 3 * sqrt 3) ..............= (sqrt 3) / 3 x = 2 +/- [(sqrt 3) / 3]
Solving by Completing the square?
Completing the square? Here's what you are asking for: X^2 - 3X - 88 = 0 Tanspose the constant to the other side X^2 - 3X = 88 Complete the squares that are missing in the (?) X^2 - 3X + (?) = 88 + (?) ( 9/4) seems the missing square X^2 - 3X + (9/4) = 88 + (9/4) Simplifying and factoring the left side of the equation .....................(4)88 + 9 (X-3/2)(X-3/2) = ------------- ..........................4 ........................361 (X-3/2)(X-3 /2) = ------ ...........................4 Extracting the square root of both sides (X- 3/2) = 19/2 X - 3/2 = 19/2 .........X = 19/2 + 3 /2 ............= 22 /2 ..........X = 11
Solve x 2 + 10x + 7 = 0 by completing the square.?
Since 5 is one half of the 10x coefficient, the square needs to be 5 squared to complete it. Add 18 to both sides to get: x^2 + 10x + 25 = 18 Factor to: (x + 5)^2 = 18 take the square roots of both sides: x + 5 = +- sqrt(18) Subtract 5 from both sides: x = -5 +- sqrt(18) Split the sqrt(18): x = -5 +- sqrt(9)sqrt(2) Simplify: x = -5 +- 3sqrt(2)
Move the constant to the right-hand side:y^2 + 12*y = -4add the square of one-half the coefficient of the y term to each side. One half of 12 is 6, so add 36y^2 + 12*y +36 = 32this becomes(y+6)^2 = 32Take square roots:y+6 = + or - sqrt(32) = + or - 4*sqrt(2)Subtract 6:y = -6 +4*sqrt(2) or y = -6–4*sqrt(2)
I am not sure what you really want to ask, but assuming you need the answer of the mentioned quadratic equation , I will provide you with its explanation.When you have a quadratic equation as ax^2+bx+c so to find the value of 'x' or so called the roots of the equation there is a very old traditional formula for this and that is calledSHRIDHARACHARYA formulawhich states that for any quadratic equation the value of its roots can be found out by x={-b+(b^2-4*a*c)^(1/2)}/2*a orx={-b-(b^2-4*a*c)^(1/2)}/2*aNow for your equation i.e 2x^2-11x+1=0a=2;b=-11;c=1put these values in the equation of the dharacharya and you will get the answer.x={-(-11)+((-11)^2-4*2*1)^(1/2)}/2*2solve this x={11+(121-8)^(1/2)}/4x={11+(113)^(1/2)}/4similarly the other root can also be found out..
Solve 8x2 – 80x = –24 by completing the square. Need help with these algebra questions?
a million) x^2 - 5x - 14 = 0 circulate the consistent. x^2 - 5x + __ = 14 + __ interior the sparkling: (b/2)^2 x^2 - 5x + 25/4 = 14 + 25/4 despite you do to a minimum of one side, you will be able to desire to do to the different to maintain it balanced. (x - __)^2 = fifty six/4 + 25/4 interior the sparkling: b/2 (x - (5/2))^2 = 80 one/4 remedy. x - (5/2) = +/- root(80 one/4) x - 5/2 = +/- (9/2) x = 9/2 + 5/2 = 14/2 = 7 x = -9/2 + 5/2 = -4/2 = -2 ~*~*~*~*~*~*~*~*~*~ 2) 4x^2 - 8x + 3 = 0 4x^2 - 8x = -3 4(x^2 - 2x + __) = -3 + __ 4(x^2 - 2x + a million) = -3 + 4 4(x - a million)^2 = a million (x - a million)^2 = a million/4 x - a million = +/- root(a million/4) x - a million = +/- (a million/2) x = a million/2 + 2/2 = 3/2 x = -a million/2 + 2/2 = a million/2
Solve the equation p2 + 4p = 1 by completing the square. Show your work.?
we need to put in the form (p + d)^2 = ? 1. expand left hand side p^2 + (2d)p + d^2 2. take your formula and add d^2 to both sides p^2 + 4p + d^2 = 1 + d^2 3. compare left-hand-side of both functions and solve for d p^2 + 4p + d^2 p^2 + (2d)p + d^2 2d = 4 d = 2 4. plug d into formula p^2 + 4p + d^2 = 1 + d^2 p^2 + 4p + 4 = 5 (p + 2)^2 = 5