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- Equation:
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Equation (x+1)^2+(x-6)^2=2x^2
- Equation sin3x+sinx-sin2x=2cos^2x-2cosx
- Equation (4+x*9)-36=40
- Equation 5/8/x=-37/48
- Express {x} in terms of y where:
- -1*x-19*y=-10
- 5*x+5*y=20
- 4*x-16*y=6
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- Integral of d{x}:
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2x^2
- Derivative of:
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2x^2
- Canonical form:
- 2x^2
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Identical expressions
- (x+ one)^ two +(x- six)^ two = two x^2
- (x plus 1) squared plus (x minus 6) squared equally 2x squared
- (x plus one) to the power of two plus (x minus six) to the power of two equally two x squared
- (x+1)2+(x-6)2=2x2
- x+12+x-62=2x2
- (x+1)²+(x-6)²=2x²
- (x+1) to the power of 2+(x-6) to the power of 2=2x to the power of 2
- x+1^2+x-6^2=2x^2
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Similar expressions
- -3x^2-5x-6=-x^2-x+(-1-2x^2)
- x^5-x^4-2x^3+2x^2-3x+3=0
- (x+1)^2-(x-6)^2=2x^2
- (x-2)^2+(x-3)^2=2x^2
- sqrt(5x^2-20x+21)+sqrt(3x^2-12x+28)=8x-2x^2-3
- 2x^2-5x-7=0
- (x+1)^2+(x+6)^2=2x^2
- (x-1)^2+(x-6)^2=2x^2
(x+1)^2+(x-6)^2=2x^2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 2 (x + 1) + (x - 6) = 2*x
$$\left(x - 6\right)^{2} + \left(x + 1\right)^{2} = 2 x^{2}$$
Detail solution
Given the equation:
Expand expressions:
Reducing, you get:
Move free summands (without x)
from left part to right part, we given:
$$- 10 x = -37$$
Divide both parts of the equation by -10
We get the answer: x = 37/10
(x+1)^2+(x-6)^2 = 2*x^2
Expand expressions:
1 + x^2 + 2*x + (x - 6)^2 = 2*x^2
1 + x^2 + 2*x + 36 + x^2 - 12*x = 2*x^2
Reducing, you get:
37 - 10*x = 0
Move free summands (without x)
from left part to right part, we given:
$$- 10 x = -37$$
Divide both parts of the equation by -10
x = -37 / (-10)
We get the answer: x = 37/10
Sum and product of roots
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sum
37 -- 10
$$\frac{37}{10}$$
=
37 -- 10
$$\frac{37}{10}$$
product
37 -- 10
$$\frac{37}{10}$$
=
37 -- 10
$$\frac{37}{10}$$
37/10
The graph