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Andrew [12]
3 years ago
14

Hi. Please I need help with these questions.

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0

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}

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BARSIC [14]
If you would like to solve the equation 3 * (3 * x - 1) + 2 * (3 - x) = 0, you can calculate this using the following steps:

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