when a polynomial p(x) is divided by 3x-1, the quotient and remainder are x^+ 2x - 3 and 5 respectively. find p(x)
Answers
Answer:
Question:
The value of a for which the given pair of equations (2a - 1)x + (a - 1)y - (2a + 1) = 0 ; 6x + 2y - 1 = 0 have no solutions is :
a = 1
a = 2
a = -1
a = -2
Answer:
Second option a = 2 is correct.
Step By step Explanation:
Given:
Two equations ; (2a - 1)x + (a - 1)y - (2a + 1) = 0 and 6x + 2y - 1 = 0
To Find:
Value of a for which given pair of equations have no solution.
Solution:
Here, we have two equations (2a - 1)x + (a - 1)y - (2a + 1) = 0 and 6x + 2y - 1 = 0, we know that in case of no solution ::
\pmb{\boxed{\bf{\purple{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}}}}
a
2
a
1
=
b
2
b
1
=
c
2
c
1
a
2
a
1
=
b
2
b
1
=
c
2
c
1
Where, \bf a_{1}a
1
is coefficient of x in first equation, \bf a_{2}a
2
is coefficient of x in second equation, \bf b_{1}b
1
is cofficient of y in first equation, \bf b_{2}b
2
is coefficient of y in second equation, \bf c_{1}c
1
is constant term in first equation and \bf c_{2}c
2
is constant term in second equation.
We have, \bf a_{1}a
1
= (2a - 1), \bf a_{2}a
2
= 6, \bf b_{1}b
1
= (a - 1), \bf b_{2}b
2
= 2, \bf c_{1}c
1
= (2a + 1) and \bf c_{2}c
2
= -1.
❏ Finding value of a :
⇒ \sf \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}
a
2
a
1
=
b
2
b
1
=
c
2
c
1
❏ Putting all known values :
⇒ \sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2} \neq \dfrac{2a + 1}{-1}
6
2a−1
=
2
a−1
=
−1
2a+1
⇒ \sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2}
6
2a−1
=
2
a−1
★ By cross multiplication :
⇒ \sf 2(2a - 1) = 6(a - 1)2(2a−1)=6(a−1)
⇒ \sf 4a - 2 = 6a - 64a−2=6a−6
⇒ \sf 6a - 4a = -2 + 66a−4a=−2+6
⇒ \sf 2a = 42a=4
⇒ \sf a = {\cancel{\dfrac{4}{2}}}a=
2
4
➠ \pmb{\blue{\underline{\boxed{\bf{\pink{a = 2}}}}}}
a=2
a=2
∴ Hence, value of a for which the given pair of equations have no solution is 2. So, second option a = 2 is correct.
Verification:
↦ \sf \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}
a
2
a
1
=
b
2
b
1
=
c
2
c
1
↦ \sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2} \neq \dfrac{2a + 1}{-1}
6
2a−1
=
2
a−1
=
−1
2a+1
❏ Putting value of a in above eqⁿ :
↦ \sf \dfrac{2(2) - 1}{6} = \dfrac{2 - 1}{2} \neq \dfrac{2(2) + 1}{-1}
6
2(2)−1
=
2
2−1
=
−1
2(2)+1
↦ \sf \dfrac{4 - 1}{6} = \dfrac{1}{2} \neq \dfrac{4 + 1}{-1}
6
4−1
=
2
1
=
−1
4+1
↦ \sf {\cancel{\dfrac{3}{6}}} = \dfrac{1}{2} \neq \dfrac{5}{-1}
6
3
=
2
1
=
−1
5
➦ \pmb{\blue{\underline{\boxed{\bf{\pink{\dfrac{1}{2} = \dfrac{1}{2} \neq -\dfrac{5}{1}}}}}}}
2
1
=
2
1
=−
1
5
2
1
=
2
1
=−
1
5
∴ Hence, Verified!
So, our ans i.e, a = 2 is correct.
★ Know More :
\clubsuit♣ In case of unique solution :
\pmb{\boxed{\bf{\red{\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}}}}}
a
2
a
1
=
b
2
b
1
a
2
a
1
=
b
2
b
1
\clubsuit♣ In case of infinitely many solutions :
\pmb{\boxed{\bf{\blue{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}}}}}
a
2
a
1
=
b
2
b
1
=
c
2
c
1
a
2
a
1
=
b
2
b
1
=
c
2
c
1
\clubsuit♣ In case of no solution :
\pmb{\boxed{\bf{\green{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}}}}\qquad\Bigg\{\pmb{\sf{Used\:above}}\Bigg\}
a
2
a
1
=
b
2
b
1
=
c
2
c
1
a
2
a
1
=
b
2
b
1
=
c
2
c
1
{
Usedabove
Usedabove}
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