the quadratic equation whose roots are 1-3^1/2 and 1+3^1/2 is

Question

ahsome · Answer

You are missing $x$ somewhere

ahsome · Answer

@gilbert936, we are missing $x$ somehwere

anonymous · Answer

i dont understand you

anonymous · Answer

you know what! can you solve it or not

ahsome · Answer

Ok. A quadratic equation has an $x^2$ inside it. In your two roots, there is no $x$, therefore, you cannot get a quadratic equation.

ahsome · Answer

In your two roots, there HAS to be an $x$ in the roots

ahsome · Answer

Or else it would just equal a number. Tell me where the $x$ is, then I can help solve it.

anonymous · Answer

you can only solve the equation using sum of roots and product of roots, try it you will surely get the answer and thanks for trying my question

ahsome · Answer

@gilbert936. We can do that, its just that you will get a number. Try it yourself:
$$1-3^{1/2} *1+3^{1/2}$$$$=(1-3^{1/2} )(1+3^{1/2})$$
Use the rule: $(a-b)(a+b)=a^2-b^2$$$(1)^2-(3^{1/2})^2$$$$=1-3$$$$=2$$
The problem is, we don't get a quadratic equation. A quadratic equation has a $x^2$ inside of it. Since we don't have a $2$ in either of the roots, there is no Quadratic equation, just a number

anonymous · Answer

your answer is wrong

anonymous · Answer

given alpha, beta as the roots of a quafratic equation, the equation is given as X^2-(alpha,beta)X + alphabeta=0      therefore alpha=1-13^1/2, beta=1+13^1/2      alpha+beta=2    alphabeta=(1-13^1/2)(1+13^1/2)=1-13=-12 the equation is X^2-2X-12