The quadratic equation in x is x2 + 2x.cos(A) + K=0. &also given summation and difference of solutions of above equation are -1 & -3 respectively. Hence find K & A?

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Answer 1 · 2018-05-05T13:40:02

$A=60^@$
$K=-2$

$x^2+2xcos(A)+K=0$

Let the solutions of the quadratic equation be $alpha$ and $beta$ .

$alpha+beta=-1$

$alpha-beta=-3$

We also know that $alpha+beta=-b/a$ of the quadratic equation.

$-1=-(2cos(A))/1$

Simplify and solve,

$2cos(A)=1$

$cos(A)=1/2$

$A=60^@$

Substitute $2cos(A)=1$ into the equation, and we get an updated quadratic equation,

$x^2+x+K=0$

Using the difference and sum of roots,

$(alpha+beta)-(alpha-beta)=(-1)-(-3)$

$2beta=2$

$beta=1$

When $beta=1$ ,

$alpha=-2$

When the roots are $1$ and $-2$ , we can get a quadratic equation as follows,

$(x-1)(x+2)$

$=x^2+x-2$

By comparison,

$K=-2$