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To complete the square, make sure all the variable terms are on one side of the equation and all the constants on the other side.
To complete the square, make sure all the variable terms are on one side of the equation and all the constants on the other side.
To complete the square, make sure all the variable terms are on one side of the equation and all the constants on the other side.
To complete the square, make sure all the variable terms are on one side of the equation and all the constants on the other side.
w = - 2 or w=- 26
x = - 1 or x=- 4
k = 17 or k=- 1
z=935 or z=65
We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with w are on one side of the equation and all constants are on the other side.
w^2+28w+52=0
⇕
w^2+28w=- 52
Next, we will add ( b2 )^2=196 to both sides of our equation. Then we will factor the trinomial on the left-hand side and solve the equation.
LHS+196=RHS+196
a^2+2ab+b^2=(a+b)^2
Add terms
sqrt(LHS)=sqrt(RHS)
Calculate root
LHS-14=RHS-14
The solutions for this equation are w=- 14 ± 12. Let's separate them into the positive and negative cases.
| w=- 14 ± 12 | |
|---|---|
| w_1=- 14 + 12 | w_2=- 14 - 12 |
| w_1=- 2 | w_2=- 26 |
We found that the solutions of the given equation are w_1=- 2 and w_2=- 26.
We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with x are on one side of the equation and all constants are on the other side.
x^2+5x+4=0
⇕
x^2+5x=- 4
Next, we will add ( b2 )^2= 254 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.
LHS+25/4=RHS+25/4
a^2+2ab+b^2=(a+b)^2
a = 4* a/4
Add fractions
sqrt(LHS)=sqrt(RHS)
sqrt(a/b)=sqrt(a)/sqrt(b)
Calculate root
LHS-5/2=RHS-5/2
The solutions for this equation are x=- 52 ± 32. Let's separate them into the positive and negative cases.
| x=- 5/2 ± 3/2 | |
|---|---|
| x_1=- 5/2 + 3/2 | x_2=- 5/2 - 3/2 |
| x_1=- 2/2 | x_2=- 8/2 |
| x_1=- 1 | x_2=- 4 |
We found that the solutions of the given equation are x_1=- 1 and x_2=- 4.
We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with k are on one side of the equation and all constants are on the other side.
k^2-16k-17=0
⇕
k^2-16k=17
b= - 16
Put minus sign in front of fraction
Calculate quotient
(- a)^2 = a^2
Next, we will add ( b2 )^2=64 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.
LHS+64=RHS+64
a^2-2ab+b^2=(a-b)^2
Add terms
sqrt(LHS)=sqrt(RHS)
Calculate root
LHS+8=RHS+8
The solutions for this equation are k=8 ± 9. Let's separate them into the positive and negative cases.
| k=8 ± 9 | |
|---|---|
| k_1=8 + 9 | k_2=8 - 9 |
| k_1=17 | k_2=- 1 |
We found that the solutions of the given equation are k_1=17 and k_2=- 1.
We want to solve the quadratic equation by completing the square. To do so, we will start by rewriting the equation so all terms with z are on one side of the equation and all constants are on the other side.
z^2-1000z+60 775=0
⇕
z^2-1000z=- 60 775
b= - 1000
Put minus sign in front of fraction
Calculate quotient
(- a)^2 = a^2
Next, we will add ( b2 )^2=250 000 to both sides of our equation. Then, we will factor the trinomial on the left-hand side and solve the equation.
LHS+250 000=RHS+250 000
a^2-2ab+b^2=(a-b)^2
Add terms
sqrt(LHS)=sqrt(RHS)
Calculate root
LHS+500=RHS+500
The solutions for this equation are z=500 ± 435. Let's separate them into the positive and negative cases.
| z=500 ± 435 | |
|---|---|
| z_1=500 + 435 | z_2=500 - 435 |
| z_1=935 | z_2=65 |
We found that the solutions of the given equation are z_1=935 and z_2=65.