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Complete 10th Class Maths Solution
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.2 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.3 Real Number
CHAPTER 2 – POLYNOMIALS – EXERCISE:-2.4
QUE :-Verify that the numbers given alongside of the cubic polynomial below are their zeroes. also verify the reltionship between the zeroes and the coeficients in each case .
(i) 2x³+x²-5x+2 :1/2,1,-2
(ii) x³-4x²+5x-2 : 2,1,1
SOL :-
(i) 2x³+x²-5x+2 :1/2,1,-2
p(x) = 2x³+x²-5x+2
zeroes for this polynomial are 1/2,1,-2
p(1/2)=2(1/2)³+(1/2)²-5(1/2)+2
=1/4+1/4-5/2+2
=1/2+2-5/2
= 0
p(1)=2(1)³+(1)²-5(1)+2
= 5-5
=0
p(-2)=2(-2)³+(-2)²-5(-2)+2
= -16+4+10+2
=0
therefore , 1/2,1,and-2 are the zeroes of the given polynomial. comparing the given polynomial with ax³+bx²+cx+d,
we obtain a=2, b=1, c=-5 ,d=2
we can take α =1/2 ,β=1, γ -2
α +β+γ =1/2+1+(-2)=-1/2=-b/a
αβ+βγ+αγ =1/2×1+1(-2)+1/2(-2)=-5/2=c/a
αβγ = 1/2×1×(-2)=(-2)/2=-d/a
therefore , the relationship between the zeroes and the coefficient is verified.
(ii) x³-4x²+5x-2 : 2,1,1
zeroes for this polynomial are 2,1,1
p(2)=2³-4(2)²+5(2)-2
= 8-16+10-2=0
p(1)=1³-4(1)²+5(1)-2
= 1-4+5-2=0
p(1)=1³-4(1)²+5(1)-2 =0
therefore , 2,1,and1 are the zeroes of the given polynomial. comparing the given polynomial with ax³+bx²+cx+d,
we obtain a=1, b=-4, c=5 ,d=-2
we can take α =2 ,β=1, γ =1
α +β+γ =2+1+(1)=4=-(-4)/1=-b/a
αβ+βγ+αγ =2×1+1(1)+2(1)=5/1=c/a
αβγ = 2×1×1=2=-(-2)/1=-d/a
therefore , the relationship between the zeroes and the coefficient is verified.
QUE :-2 Find a cubic polynomial with the sum ,sum of the product of its zeroes taken two at a time and the product of its zeroes as 2,-7,-14respectively.
SOL:-
Let polynomial be ax³+bx²+cx+d and the zeroes be α,β,andγ
it is given that
α +β+γ =2/1=-b/a
αβ+βγ+αγ =-7/1=c/a
αβγ = -14/1=-d/a
if a- 1 b=-2,c=-7 d=14
hence,the polynomial is x³-2x²-7x+14.
QUE:-3 If the zeroes of polynomail , x³-3x²+x+1 are a-b,a,a+b find a and b.
SOL;-
p(x) = x³-3x²+x+1
zeroes are a-b , a,a+b
comparing the given polynomial with px³+qx²+rx+t, we obtain p=1,q=-3,r=1 ,t=1
sum of zeroes =a-b+a+a+b
-q/p =3a
-(-3)/1=3a
a=1
the zeroes are 1-b ,1,1+b
multipication of zeroes = 1(1-b)(1+b)
-t/p = 1-b²
-1/1=1-b²
b² =1+1
b=±root2
hence , a=1 and b=root2 or -root2
QUE :-4 It two zeroes ofthe polynomial ,x (power)4 -6x³-26x²+138x-35 are 2±root 3 find other zeroes.
SOL:-
given 2+root 3 and 2-root 3 are zeroesof the given polynomial.
so ,(2+root 3)(2-root 3) is a factor of polynomial .
therefore , [x-(2+root 3)][x-(2-root 3)]=x² +4 – 4x-3
=x² – 4x+1 is a factore of the given polynomial
for finding the remaining zeroes of the given polynomial , we will find
x² – 4x+1 ) x (power)4 -6x³-26x²+138x-35 (x² -2x -35
x (power)4 -4x³+ x²
– + –
-2x³ -27x² +138x -35
-2x³+8x² -2x
+ – +
-35x² +140x -35
-35x² +140x -35
+ – +
0
x (power)4 -6x³-26x²+138x-35 = ( x² – 4x+1)(x² -2x -35)
( x² -2x -35) is also a factor of the given
it can be observed that polynomial (x² -2x -35) =(x-7)(x+5)
therefore , the value of the polynomial is also zero when x-7 =0 or x+5 or x=7 or -5
hence ,7 and -5 are also zeroes of this polynomial.
QUE :-5 If the polynomial x(power4) -6x³ +16x² -25x +10 is dividend by another polynomial , x²-2x+k the remainder comes out to be comes out to be x+a , find k and a.
SOL:- by division algorithm ,
Dividend = divisor × quotient + remainder
Dividend – remainder = divisor × quotient
x(power4) -6x³ +16x² -25x +10 -x-a =x(power4) -6x³ +16x² -26x +10-a
x(power4) -6x³ +16x² -26x +10-a divisible by x²-2x+k
x²-2x+k ) x(power4) -6x³ +16x² -26x +10-a (x² -4x +(8-k)
x(power4) -2x³ +kx²
– + –
-4x³ +(16-k) x² -26 x
-4x³ + 8x² -4kx
+ – +
(8-k)x² -(26-4k)x +10-a
(8-k )x² -(16 -2k) + (8k-k²)
– + –
(10 +2k)x +(10-a -8k +k² )
it can be observed that (10 +2k)x +(10-a -8k +k² ) will be 0.
therefore (-10 +2k) =0 and (10-a -8k +k² ) =0
for (-10 +2k) =0 ,2k=10 and k=5
for (10-a -8k +k² ) =0
10-a-8×5+25 =0
10-a-40+25=0
-5-a =0
therefore ,a=-5
hence , k=5 and a=-5
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