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QUADRATIC EQUATIONS
The polynomial of degree two is called quadratic polynomial and equation corresponding to a quadratic polynomial P(x) is called a quadratic equation in variable x.
Thus, P(x) = ax2 + bx + c =0, a ≠ 0, a, b, c ∈ R is known as the standard form of quadratic equation.
There are two types of quadratic equation.
(i) Complete quadratic equation : The equation ax2 + bx + c 0 where a ≠ 0, b ≠ 0,c ≠ 0
(ii) Pure quadratic equation : An equation in the form of ax2 = 0, a ≠ 0, b = 0, c = 0
Posted by Khushi Singh 16 hours ago
- 1 answers
Sia 🤖 16 hours ago
Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.
Posted by Sarita Agarawal 1 day, 6 hours ago
- 1 answers
Rajan Kumar Pasi 1 day, 1 hour ago
{tex}Question:\space\large{ {2x-3\over x-2} +{2x-7\over x-4} = {16\over 3}, find \space x?}{/tex}
{tex}\large {Solution:\\ } {/tex}
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{tex}\large { \underline {Step\space 1:\space Fraction\space solving} } {/tex} {tex}\large\implies {{\big((2x-3)(x-4)\big)} +{\big((2x-7)(x-2)\big)}\over {(x-2)(x-4)}} = {16\over 3}{/tex} {tex}\large\implies {{\big((2x^2-8x-3x+12)\big)} +{\big((2x^2-4x-7x+14)\big)}\over {(x^2-4x-2x+8)}} = {16\over 3}{/tex} {tex}\large\implies {{\big(2x^2-11x+12+2x^2-11x+14\big)}\over {(x^2-6x+8)}} = {16\over 3}{/tex} {tex}\large\implies {{\big(4x^2-22x+26\big)}\over {(x^2-6x+8)}} = {16\over 3}{/tex} |
{tex}\large{ \underline {Step\space 2:\space Cross-multiplication}\\\implies {{3\times \big (4x^2-22x+26\big)} } = {16\times {(x^2-6x+8)}}}{/tex}
{tex}\large{ \underline {Step\space 3:\space Simplification:}\\\implies {{\big (12x^2-66x+78\big)} } = {{(16x^2-96x+128)}}}{/tex}
{tex}\large {\underline {Step\space 4:\space Equation-solving:}\\\implies 16x^2-96x+128-12x^2+66x-78=0}{/tex} |
{tex}\large {\underline {Step\space 5:\space Arranging\space like-terms:}\\\implies 16x^2-12x^2-96x+66x+128-78=0}{/tex} {tex}\large{\\\implies 4x^2-30x+50=0}{/tex}
{tex}\large {\underline {Step\space 6:\space Taking\space out\space common-terms:}\\\implies 2\times (2x^2-15x+25)=0}{/tex} {tex}\large {\\\implies 2x^2-15x+25=0}{/tex} |
{tex}\large {\underline {Step\space 7:\space Middle-term\space splitting\space :}\\\implies 2x^2-10x-5x+25=0}{/tex} {tex}\large{ \underline {Step\space 8:\space Grouping\space :}\\\implies 2x(x-5)-5(x-5)=0}{/tex} {tex} \\\implies \large(2x-5)(x-5)=0{/tex} {tex} \\\implies\huge \therefore \boxed{x=5\space or \space {5\over2}}{/tex} |
Posted by Riya Singh 1 day, 7 hours ago
- 2 answers
Saraswati Sabat 18 hours ago
Rajan Kumar Pasi 1 day, 1 hour ago
This is a very simple question of linear equations in two variables chapter, I think you have to practice more on fractions arithmetic.
{tex}\large Question:\frac 2x +\frac 3{3y}=\frac 16\space and\space \frac 3x +\frac 2y=0,\space find\space x,y?{/tex}
Solution:
<th colspan="2" scope="col">|
{tex}\large\implies\boxed {\frac {2\times 3y +3\times x}{x\times 3y}=\frac 16 \space \dots eq.(i)}{/tex} {tex}\large{\implies {\frac {6y +3x}{3xy}=\frac 16 \space }}{/tex} {tex}\large {\underline {Step\space 2:\space Cross-multiplication:} }{/tex} {tex}\large{\implies {6\times{(6y +3x)}=1\times {3xy} \space }}{/tex} {tex}\large{\implies {36y +18x}={3xy} \space }{/tex} {tex}\large{\implies {36y +18x}-{3xy}=0 \space }{/tex} |
{tex}\large {\implies\boxed {\frac {3\times y +2\times x}{x\times y}=0 \space\dots eq.(ii) }}{/tex} {tex}\large {\implies\frac {3y +2x}{xy}=0 \space }{/tex} {tex}\large {\underline {Step\space 2:\space Cross-multiplication:} }{/tex} {tex}\large {\implies{3y +2x}=0\times {xy}=0 \space }{/tex} {tex}\large {\implies\boxed {{3y}=-2x \space \dots eq.(iii)}}{/tex}
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{tex}\large {\underline {Step\space 3:\space Putting\space relation\space of\space eq.(iii):} }{/tex} {tex}\large{\implies {{\underline {3y}}(12-x) +18x}=0 \space }{/tex} {tex}\large{\implies {\underline{(-2x)}(12-x) +18x}=0 \space }{/tex} {tex}\large{\implies {-24x+2x^2 +18x}=0 \space }{/tex} {tex}\large{\implies {2x^2-6x}=0 \space }{/tex} {tex}\large{\implies {2x^2=6x \space} }{/tex} {tex}\large{\implies\boxed {x=2 \space} }{/tex} |
{tex}\large {\underline {Step\space 4:\space Putting\space value\space of\space 'x'\space in\space eq.(iii):} }{/tex}{tex}\large {\implies{3y}=-2x \space }\\ \large {\implies{3y}=-2\times 2 \space }\\ \large\implies\boxed {{y}={-4\over3} \space }\\{/tex} |
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Posted by Danish Roshan 1 day, 8 hours ago
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Rajan Kumar Pasi 1 day ago
This also seems like you didn't learn chapter 4 in a better way.
{tex}\large{Question: \space Zeroes=3+\sqrt5 \space and \space 3-\sqrt 5,\space find\space the \space quadratic\space equation?}{/tex}
{tex}\large Solution:{/tex}
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{tex}\large {For\space any\space quadratic\space equation,\space x^2+x\times(\alpha+\beta)+(\alpha\times\beta)=0\space }{/tex} {tex}\large {Here,\space let\space\space\alpha=3+\sqrt5 \space\space\space\space,\space\space\space\space\beta=3-\sqrt5.\space }{/tex} {tex}\large {Thus,\space\space\space\alpha+\beta=3+\sqrt5 +3-\sqrt5 = 6\space\space\space\space,\space\space and\space\space\alpha\times\beta=(3+\sqrt5)(3-\sqrt5)=3^2-\sqrt5 ^2=9-5=4.\space }{/tex} {tex}\large {\therefore\space x^2+x\times(\alpha+\beta)+(\alpha\times\beta)=0\space }\\ \large {\implies\space x^2+x\times(6)+(4)=0\space }\\ \large {\implies\boxed{\space x^2+6x+4=0}\space }\\{/tex} |
Now after seeing this, don't you think this is a very very simple thing to do.
Posted by Dalu Chowdhury 1 day, 8 hours ago
- 1 answers
Rajan Kumar Pasi 1 day ago
Question : Can 72 and 20 be the L.C.M and H.C.F of two numbers? Write down the reason.
This is a very pathetic question to ask. It really shows that you are not studying well.
Solution: {tex}\large{\implies \boxed{LCM\times HCF=a\times b}}{/tex}
{tex}\large{\implies{72\times20=a\times b}}{/tex} {tex}\large{\implies{1440=a\times b}}{/tex}
Now, 1440 can be multiple of any two numbers.
{tex}\large{\implies{\space\space1\times1440=1440 }}\\ \large{\implies{\space\space2\times720\space\space=1440 }}\\ \large{\implies{\space\space3\times480\space\space=1440 }}\\ \large{\implies{\space\space4\times360\space\space=1440}}\\ \large{\implies{\space\space5\times288\space\space=1440}}\\ \large{\implies{\space\space6\times240\space\space=1440}}\\ \large{\implies{\space\space8\times180\space\space=1440}}\\ \large{\implies{\space\space9\times160\space\space=1440 }}\\ \large{\implies{10\times144\space\space=1440 }}\\ \large{\implies{12\times120\space\space=1440 }}\\ \large{\implies{15\times96\space\space\space\space=1440 }}\\ \large{\implies{16\times90\space\space\space\space=1440 }}\\ \large{\implies{18\times80\space\space\space\space=1440 }}\\ \large{\implies{20\times72\space\space\space\space=1440 }}\\ \large{\implies{24\times60\space\space\space\space=1440 }}\\ \large{\implies{30\times48\space\space\space\space=1440 }}\\ \large{\implies{32\times45\space\space\space\space=1440 }}\\ \large{\implies{36\times40\space\space\space\space=1440 }}\\ {/tex}
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