Question

If $\alpha ,\beta$ are roots of the equation ${2x}^2-35x+2=0$ than evaluate ${(2\alpha -35)}^2$+ ${(2\beta -35)}^2$

Answer 1

We have,

$2x^2-35x+2=0$

On comparing that,

$A{x}^2+Bx+C=0$

Then,

$A=2,B=-35,C=2$

Then,

Using quadratic formula,

$x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

Then,

$x=\frac{35\pm \sqrt{{35}^2-4\times 2\times 2}}{2\times 2}$

$=\frac{35\pm \sqrt{1225-16}}{4}$

$=\frac{35\pm \sqrt{1209}}{4}$

$=\frac{35\pm 34.77}{4}$

Now,

$\alpha =\frac{35+34.77}{2}$

$\alpha =\frac{69.77}{2}=34.885$

$\alpha =34.885$

$\beta =\frac{35-34.77}{2}$

$\beta =\frac{0.23}{2}$

$\beta =0.115$

Then,

Calculate

${(2\alpha -35)}^2+{(2\beta -35)}^2$

$\Rightarrow {(2\times 34.885-35)}^2+{(2\times 0.115-35)}^2$

$\Rightarrow 1208.9529+1208.9529$

$\Rightarrow 2417.9058$

Hence, this is the answer.