Question

Hi. Please I need help with these questions.
It's urgent . 50pts.
No jokes. This time
Answer No 12 and 13

Answer 1

Answer:

Q 12 roots of the equation

$2x^{2} -6+1=0 \\= (x-\frac{\sqrt{10} }{2})(x+\frac{\sqrt{10} }{2})$

∝ = $\frac{\sqrt{10} }{2}$

β = $-\frac{\sqrt{10} }{2}$

no matter if u oppose the root

(i) 2($\frac{\sqrt{10} }{2}$)$(-\frac{\sqrt{10} }{2} )^{2}$+2$(\frac{\sqrt{10} }{2} )^{2}$($-\frac{\sqrt{10} }{2}$)+2(

(ii)($(\frac{\sqrt{10} }{2})^{2}$ - 3 ($\frac{\sqrt{10} }{2}$)($-\frac{\sqrt{10} }{2}$) + ($(-\frac{\sqrt{10} }{2})^{2}$) = $\frac{25}{2}$

Q 13  roots of equation

$4x^{2} -3x-4=0\\\alpha = -0.693\\\beta = 1.443$

the roots of the second equation are

x1 = 1/3(-0.693) = -0.231

x2 = 1/3(1.443) = 0.481

the equation is

(x+0.231)(x-0.481)=0

$x^{2}-\frac{1}{4} x-\frac{1}{9}$