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Sia ? 6 years, 4 months ago
The polynomial with α and β as roots is:
f(x)=x2-(α+β)x+αβ
x2-{tex}(\frac 23-\frac 14){/tex}x+{tex}(\frac 23)(-\frac 14){/tex}
{tex}\mathrm x^2-\frac5{12}\mathrm x-\frac16{/tex}
or f(x)=12x2-5x-2
here a=12,b=-5,c=-2
Sum of zeros = {tex}\frac23-\frac14=\frac5{12}=-\frac{\mathrm b}{\mathrm a}{/tex}
Product of zeros = {tex}\frac23\times-\frac14=-\frac16=\frac{-2}{12}=\frac{\mathrm c}{\mathrm a}{/tex}
Hence relation of coefficients and zeros is verified
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{tex}\frac { 1 } { x + 4 } - \frac { 1 } { x - 7 } = \frac { 11 } { 30 }{/tex} where {tex}x \neq - 4,7{/tex}
{tex}\Rightarrow \frac { ( x - 7 ) - ( x + 4 ) } { ( x + 4 ) ( x - 7 ) } = \frac { 11 } { 30 }{/tex}
{tex}\Rightarrow \frac { - 11 } { ( x + 4 ) ( x - 7 ) } = \frac { 11 } { 30 }{/tex}
{tex}\Rightarrow{/tex} x2 - 7x + 4x - 28 = -30
{tex}\Rightarrow{/tex} x2 - 3x + 2= 0
Comparing equation x2 - 3x + 2 = 0 with general form ax2 + bx + c = 0,
We get a = 1, b = -3 and c = 2
Using quadratic formula {tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}to solve equation,
{tex}x = \frac { 3 \pm \sqrt { ( - 3 ) ^ { 2 } - 4 ( 1 ) ( 2 ) } } { 2 \times 1 }{/tex}
{tex}\Rightarrow x = \frac { 3 \pm \sqrt { 1 } } { 2 }{/tex}
{tex}\Rightarrow x = \frac { 3 + \sqrt { 1 } } { 2 } , \frac { 3 - \sqrt { 1 } } { 2 }{/tex} {tex}\Rightarrow{/tex} x = 2, 1
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Milind Patel 7 years, 7 months ago
1Thank You