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Yogita Ingle 6 years, 9 months ago
x + y =5 ... (i)
2x –3y = 4 ... (ii)
Multiplying equation (i) by 2, we get
2x + 2y = 10 ... (iii)
2x –3y = 4 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 - (6/5) = 19/5
Hence, x = 19/5 and y = 6/5
Posted by Sarma Schwaantech 6 years, 9 months ago
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Ravi.....✌️ Ivar......? 6 years, 9 months ago
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Posted by Nandini Sharma❤️ 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
Let p(x) = x3 + ax2 + bx + c
As -1 is one of the zeroes of p(x), p(-1) = 0
⇒ (-1)3 + a(-1)2 + b(-1) + c = 0
⇒ - 1 + a – b + c = 0
⇒ c = b –a + 1 …. (i)
Let {tex}\alpha{/tex}, β be two other zeroes of p(x),
then product of zeroes =-1× {tex}\alpha{/tex} × β ={tex}\;\;-\frac{\;\;Cons\tan t\;term\;}{coefficient\;of\;x^3}{/tex}
⇒ (-1) (α β) = {tex}\;\;-\frac{\;\;c}1{/tex}
⇒ - {tex}\alpha{/tex} β = - c
⇒ {tex}\alpha{/tex} β = c
⇒ {tex}\alpha{/tex} β = b –a + 1 [using (i)]
Hence, the product of other two zeroes of the given cubic polynomial is b – a + 1.
Posted by Nandini Sharma❤️ 6 years, 9 months ago
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Sia ? 6 years, 7 months ago
Given:
f(x) = (2x4 – 9x3 + 5x2 + 3x – 1)
Zeroes = (2 + √3) and (2 – √3)
Given the zeroes, we can write the factors = (x – 2 + √3) and (x – 2 – √3)
{Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}
Multiplying these two factors, we can get another factor which is:
((x – 2) + √3)((x – 2) – √3) = (x – 2)2 – (√3)2
⇒x2 + 4 – 4x – 3 = x2 – 4x + 1
So, dividing f(x) with (x2 – 4x + 1)
f(x) = (x2 – 4x + 1) (2x2 – x – 1)
Solving (2x2 – x – 1), we get the two remaining roots as
{tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}
where f(x) = ax2 + bx + c = 0(using Quadratic Formula)
{tex}\mathrm{x}=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(2)(-1)}}{2(2)}{/tex}
{tex}\mathrm{x}=\frac{-1 \pm 3}{4}{/tex}
{tex}\Rightarrow \mathrm{x}=1,-\frac{1}{2}{/tex}
Zeros of the polynomial = {tex}1,-\frac{1}{2}, 2+\sqrt{3}, 2-\sqrt{3}{/tex}
Posted by Jay Joshi 6 years, 9 months ago
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Ravi.....✌️ Ivar......? 6 years, 9 months ago
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