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Simplify 3 cubed radical 128?
3^3*radical(128) = 3^3*radical(2^7) = 3^3*radical(2^6*2) =3^3*2^3*radical(2) = 216*radical(2).
What is -3 radical 2 radical 50?
-3*radical(2)*radical(50) = -3*radical(2*50) = -3*radical(100) = -3*10 = -30
What is radical 14 times radical 2?
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.
What are examples of quadratic formula with 2 imaginary roots?
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.
What is the equation of a quadratic formula that has answers of -4 and 3?
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
