Date :if the zeros of polynomial ax^2+bx+carealpha and ß then find the zerosof polynomial cx^2 – 2bx+ 4a including& and ß About the author Daisy
Step-by-step explanation: Let, z be the general zero of ax²+bx+c i.e. [tex]z = \alpha \: or \beta [/tex] Then, az²+bz+c=0 Let, [tex]z = – \frac{2}{y} [/tex] Then, [tex]a( – \frac{2}{y})^{2} + b( – \frac{2}{y}) + c = 0 [/tex] Multiplying both side by y² [tex]4a – 2by + cy^{2} = 0[/tex] [tex] = cy^{2} – 2by + 4a = 0[/tex] But this is the equation whose roots we are looking for. We observe that y is the general solution of this equation. But y= -2/z But [tex]z = \alpha \: or \beta [/tex] Therefore [tex]y = – \frac{2}{ \alpha } \: or \: – \frac{2}{ \ \beta }[/tex] This is the required solution. Reply
Step-by-step explanation:
Let,
z be the general zero of ax²+bx+c i.e.
[tex]z = \alpha \: or \beta [/tex]
Then,
az²+bz+c=0
Let,
[tex]z = – \frac{2}{y} [/tex]
Then,
[tex]a( – \frac{2}{y})^{2} + b( – \frac{2}{y}) + c = 0 [/tex]
Multiplying both side by y²
[tex]4a – 2by + cy^{2} = 0[/tex]
[tex] = cy^{2} – 2by + 4a = 0[/tex]
But this is the equation whose roots we are looking for.
We observe that y is the general solution of this equation.
But y= -2/z
But
[tex]z = \alpha \: or \beta [/tex]
Therefore
[tex]y = – \frac{2}{ \alpha } \: or \: – \frac{2}{ \ \beta }[/tex]
This is the required solution.