3^3*radical(128) = 3^3*radical(2^7) = 3^3*radical(2^6*2) =3^3*2^3*radical(2) = 216*radical(2).
-3*radical(2)*radical(50) = -3*radical(2*50) = -3*radical(100) = -3*10 = -30
Well, honey, radical 14 times radical 2 is just radical 28. It's like multiplying two annoying siblings who always want attention - they combine to become one big radical mess. So, there you have it, radical 28 is the result of that math family reunion.
A quadratic equation has the form: x^2 - (sum of the roots)x + (product of the roots) = 0 If the roots are imaginary roots, these roots are complex number a+bi and its conjugate a - bi, where a is the real part and b is the imaginary part of the complex number. Their sum is: a + bi + a - bi = 2a Their product is: (a + bi)(a - bi) = a^2 + b^2 Thus the equation will be in the form: x^2 - 2a(x) + a^2 + b^2 = 0 or, x^2 - 2(real part)x + [(real part)^2 + (imaginary part)^2]= 0 For example if the roots are 3 + 5i and 3 - 5i, the equation will be: x^2 - 2(3)x + 3^2 + 5^2 = 0 x^2 - 6x + 34 = 0 where, a = 1, b = -6, and c = 34. Look at the denominator of this quadratic equation: D = b^2 - 4ac. D = (-6)^2 - (4)(1)(34) = 36 - 136 = -100 D < 0 Since D < 0 this equation has two imaginary roots.
The roots of the equation are -4 and 3. Simplest solution is (x +4 )(x - 3) which multiplies out as x^2 + x - 12 = 0
A quadratic equation has two roots. They may be similar or dissimilar. As the highest power of a quadratic equation is 2 , there are 2 roots. Similarly, in the cubic equation, the highest power is 3, so it has three equal or unequal roots. So the highest power of an equation is the answer to the no of roots of that particular equation.
3^3*radical(128) = 3^3*radical(2^7) = 3^3*radical(2^6*2) =3^3*2^3*radical(2) = 216*radical(2).
roots of equation are x values when y = 0
-3*radical(2)*radical(50) = -3*radical(2*50) = -3*radical(100) = -3*10 = -30
Well, honey, radical 14 times radical 2 is just radical 28. It's like multiplying two annoying siblings who always want attention - they combine to become one big radical mess. So, there you have it, radical 28 is the result of that math family reunion.
It is a quadratic equation and when solved it has equal roots of 3/2 or 1.5
One possibility is 7/2 + sqrt(1/4)
The equation x2+5x+6=0 simplifies to (x+2)*(x+3)=0. From this you can determine the roots by setting x+2 and x+3 equal to zero. The roots of the equation are -2 and -3.
It is a quadratic equation and can be rearranged in the form of:- x2-x-6 = 0 (x+2)(x-3) = 0 Solutions: x = -2 and x = 3
Use the quadratic equation to find the roots of 2x2-3x-3=0.a=2, b=-3, c=-3x =[ -b +- SQRT(b2-4ac)]/2a=[--3 +- SQRT((-3)2 - 4(2)(-3))]/2*2= [3+-SQRT(9--24)]/4so the roots are x = [3+SQRT(33)]/4 and x = [3-SQRT(33)]/4
There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2
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