प्रश्नावली-3A
1. निम्नलिखित समीकरणों में द्विघात समीकरण को पहचानें|
(i) 6+4x=0
(ii) ax2+bx+c=0, where a=0
(iii) x(x+3)=x2
(iv) (x-2) (x+2)=5
(v) x2 + 1/x2 =2 , x=/0
(vi) x2=x+3/x
(vii) x2-4×2-x+2
(viii) (x+1)2=2(x-2)
(ix) x(x+1)+7=(x+2) (x-2)
(x) (x-2)2+2=2x-3
Answer:- (iv), (vii), (viii), (x)
व्याख्या:-
(i)नहीं क्योंकि चर का x घात 2 नहीं है|
(ii)नहीं क्योंकि a=0 है|
(iii)नहीं क्योंकि x2=x2 कट जाएगा|
(v)नहीं क्योंकि भिन्न का मान करने पर x होगा|
(vi)नहीं क्योंकि x2=x2
(ix) नहीं क्योंकि x2=x2
2. (i) क्या 2×2-5x+2=0 का एक मूल 5 होगा क्यों?
उत्तर:-
2×(5)2-5×5+2=0
50-25+2=27
समीकरण में 5 मान रखने पर 0 नहीं आया इसिलिए नहीं होगा|
(ii) जांच कीजिये कि x=1 समीकरण 3×2-2x-1=0 का हल है या नहीं|
Answer:- 3×2-2x-1=0
3×(1)2-2×1-1=0
3-2-1=0
Yes
(iii) क्या x=-1/2,2/3 समीकरण 6×2-x-2=0 के मूल है क्यों?
Answer:-
6×2-x-2=0
x=-1/2, 6×( -1 )2- ( -1 ) -2=0
2 2
6 + 1 -2=0
4 2
6+2-8 =0
4
8-8 =0
4
0/4=0
x=2/3, 6×( 2 )2- 2 -2=0
3 3
6×4 – 2 -2=0
9 3
24-6-18 =0
9
24-24 =0
9
0/9=0
(iv) क्या x=√2,-2√2 समीकरण x2+√2x-4=0 के मूल है क्यों?
Answer:-
x2+√2x-4=0
x=√2, (√2)2+√2×√2-4=0
2+2-4=0
x=-2√2, (-2√2)2+√2×(-2√2)-4=0
8-4-4=0
3. (x-2) (x+2) =4
Answer:-
(x-2) (x+2) =4
x×x+2x-2x-4=4
x2-4=4
x2=4+4=8
x=√8
x=2√2, -2√2
4. x2=9
Answer:-
x2=9
x=√9
x=3, -3
5. समीकरण (x+2) (3x-1)=0 का हल समुच्चय लिखिये
उत्तर:-
(x+2)(3x-1)=0
x×3x-x×1+2×3x-2×1=0
3×2-x+6x-2=0
3×2+5x-2=0
D=b2-4ac= (5)2-4•3•-(2)
25+24=49
x= -b+-√b2-4ac = -b+-√D
2a 2a
= -5 + – √49 = -5+-7
2×3 6
= -5-7 -5+7
6 6
= -12 =-2 2 =1/3
6 6
6. रिक्त स्थान को भरिए——
समीकरण x2-5=0
x2=5
7. निम्नलिखित द्विघात समीकरणों में कौन से समीकरण का हल परिमेय संख्याओं के समुच्चय में होगा|
उत्तर:-
x2=5
x=√5
x(x-2)=3-2x
x2-2x=3-2x
x2-2x+2x=3
x2=3
x=√3
(x+1)(x-1)=0
x2+x-x-1=0
x2-1=0
x2=1
x=√1=1, -1
8. समीकरण का केवल मूल्य समुच्चय लिखें-
उत्तर:-
(i) 4×2=1
x2=1/4
x=√ 1
4
x=+- 1
2
(ii) 7 x2=35
5
7×2=35×5
x2= 35×5
7
x2=5×5
x=5, -5
(iii) 9×2+2=18
9×2=18-2=16
x2= 16
9
x= √ 16
9
x= 4 , -4
3 3
(iv) 11×2-8=8-14×2
11×2+14×2=8+8=16
25×2=16
x2= 16
25
x= √ 16
25
x= 4 , -4
5 5
10. (I) समीकरण 6m2-7m+2=0 का हल समुच्चय निकालें-
(ii) x2-2x+1=1
Answer:-
(i) 6m2-7m+2=0
6m2-4m-3m+2=0
2m(3m-2)-1(3m-2)=0
(3m-2)(2m-1)=0
3m-2=0 2m-1=0
3m=2 2m=1
m=2/3 m=1/2
(ii) x2-2x+1=1
x2-2x+1-1=0
x2-2x=0
x(x-2) =0
x=0, x-2=0, x=2
11. द्विघात सूत्र से हल करें—–
(i) 3×2-5x+2=0
a=3, b=-5, c=2
D=b2-4ac=(-5)2-4×3×2=25-24=1
X=-b+-√D =-(-5)+-√1 = 5+-1
2a 2×3 6
= 5+1 5-1
6 6
= 6 =1 4 = 2
6 6 3
(ii) 2×2-2√2x+1=0
a=2, b=-2√2, c=1
D=b2-4ac=(-2√2)2-4×2×1=8-8=0
X= -b+-√D = -(-2√2)+-√0 = 2√2
2a 2×2 4
=1/√2 and 1/√2
12. Solve:- x2-3x-10=0
x2-5x+2x-10=0
x(x-5)+2(x-5)=0
(x-5)(x+2)=0
x-5=0 x+2=0
x=5 x=-2
13. (i) 2×2-5x+3=0
(ii) ax2-2abx=0
Answer:-(i) 2×2-5x+3=0
a=2, b=-5, c=3
D=b2-4ac=(-5)2-4×2×3=25-24=1
D=-b+-√D = -(-5)+-√1 = 5+- 1
2a 2×2 4
5+1 5-1
4 4
6 4 =1
4 4
3
2
(ii) ax2-2abx=0
a=a, b=-2ab, c=0
D=b2-4ac=(-2ab)2-4×a×0=4a2b2
X= -b + – √D = -(-2ab) +- √4a2b2
2a 2×a
2ab+-2ab = 4ab =2b
2a 2a
2ab-2ab = 0/2a=0
2a
14. निम्नलिखित द्विघात समीकरणों को हल करें-
उत्तर:-
(i) 4 x2-2x+ 3 =0
3 4
a=4/3, b=2, c=3/4
D=b2-4ac=(2)2-4× 4 × 3 =4-4=0
3 4
X=-b+-√D = -(-2)+-√0 = 2 = 2×3
2a 2×4/3 8/3 8
=3/4, 3/4
(ii) 2z2+z-6=0
a=2, b=1, c=-6
D=b2-4ac=(1)-4×2×(-6)=1+48=49
X=-b+-√D = -1+-√49 = -1+- 7
2a 2×2 4
-1-7 -1+7
4
-8 =-2 6 = 3
4 4 2
(iii) 3y2-2√6y+2=0
a=3, b=-2√6, c=2
D=b2-4ac=(-2√6)2-4×3×2=24-24=0
X= -b+-√D = -(-2√6)+-√0 = 2√6
2a 2×3 6
2√6 = √2×√2×√6 = √2 , √2
√6×√6 √3×√2×√6 √3 √3
(iv) 2y2-y+ 1 =0
8
a=2, b=-1, c=1/8
D=b2-4ac=(-1)2-4×2× 1 =1-1=0
8
X= -b+-√D = -(1) +-√0 = 1 , 1
2a 2×2 4 4
15. Solve:-
(i) (x+3)(x-3)=7
x2-3×3=7
x2-9=7
x2=7+9=16
x2=16
x=√16=4, -4
(ii) (4x-1)(x+2)=2+7x
(4x)×(x)+8x-x-2=2+7x
4×2+7x-2=2+7x
4×2-7x-2-2-7x=0
4×2-4=0
4×2=4
x2=4/4=1
x=√1=1, -1
(iii) (2x-1)2=(x-1)(x-3)+1
(2x)2-2•2×1+1=x×x-3x-x+3+1
4×2-4x+1=x2-4x+4
4×2-4x+1-x2+4x-4=0
3×2-3=0
3x(x-1)=0
3x=0 x-1=0
x=0 x=1
16. Solve:-
(i) 2x- 3 =5
x
2×2-3=5x
2×2-5x-3=0
2×2-6x+x-3=0
2x(x-3)+1(x-3)=0
(x-3)(2x+1)=0
x-3=0 2x+1=0
x=3 2x=-1, x=-1/2
(ii) x + 2 = x + 8
2 x 8 x
x2+4 = x2+64
2x 8x
8×3+32x=4×3+128x
8×3-4×3+32x-128x=0
4×3-96x=0
4×3=96x
4×2=96
x2=96/4=16
x2=√16
x=4, -4
(iii) 2x = 5+3x
5-3x 8x
2x×8x=5×5-3x×3x
16×2=25-9×2
16×2+9×2=25
25×2=25
x2=25/25=1
x2=1
x=√1=1, -1
(iv) x+3 = 3x-7
x+2 2x-3
(x+3)(2x-3)=(x+2)(3x-7)
x×2x-3×x+3×2x-3×3=x×3x-7×x+2×3x-7×2
2×2-3x+6x-9=3×2-7x+6x-14
2×2-3x+6x-9-3×2+7x-6x+14=0
-x2+4x+5=0
x2-4x-5=0
x2+5x-x-5=0
x(x+5)-1(x+5)=0
(x+5)(x-1)=0
x+5=0 x-1=0
x=-5 x=1
(v) z+ 1 =3
z
z2+1=3z
z2-3z+1=0
a=1, b=-3, c=1
D=b2-4ac=(-3)2-4•1•1=9-4=5
x= -b+-√D = -(-3)+-√5
2a 2×1
3+- √5 = 3+√5 , 3-√5
2 2 2
(vi) 1 – 1 =3
y y-2
y-2-y =3
y(y-2)
-2 =3
y2-2y
3y2-6y=-2
3y2-6y+2=0
a=3, b=-6, c=2
D=b2-4ac=(-6)2-4•3•2=36-24=12
x= -b+- √D = -(-6)+-√12
2a 2×3
6+- √12 = 6+- 2√3 = 2(3+-√3)
6 6 6
3+-√3 = 3+√3 , 3-√3
3 3 3
17. सूत्र S=t2+t में t के किस मान के लिए s, 2 के बराबर है|
Answer:-
t=1
S=(1)2+1=1+1=2
t=-2
s=(-2)2-2=4-2=2
18. (i) 1 + 1 = 3
x+6 x-10 x-4
x-10+x+6 = 3
(x+6)(x-10) x-4
2x-4 = 3
x2-10x+6x-60 x-4
3×2-12x-180=2×2-4x-8x+16
3×2-12x-180-2×2+4x+8x-16=0
x2-196=0
x2=196
x=√196=14, -14
(ii) 1 – 1 = 1
x-3 x+5 6
x+5-x+3 = 1
(x-3)(x+5) 6
8 = 1
x2+5x-3x-15 6
x2+2x-15=48
x2+2x-15-48=0
x2+2x-63=0
x2+9x-7x-63=0
x(x+9)-7(x+9)=0
(x+9)(x-7)=0
x+9=0 x-7=0
x=-9 x=7
(iii) x + 3 = 2(6+x)
3 6-x 15
x(6-x)+3×3 = 12+2x
3(6-x) 15
6x-x2+9 = 12+2x
18-3x 15
90x-15×2+135=(12+2x) (18-3x)
90x-15×2+135=216-36x+36x-6×2
90x-15×2+135=216-6×2
90x-15×2+135-216+6×2=0
-9×2+90x-81=0
-9(x2-10x+9) =0
x2-10x+9=0
x2-9x-x+9=0
x(x-9)-1(x-9)=0
(x-9)(x-1)=0
x-9=0 x-1=0
x=9 x=1
(iv) x-1 + x-3 = 10
x-2 x-4 3
x2-4x -x+4+x2-3x-2x+6 = 10
(x-2) (x-4) 3
2×2-5x+4-5x+6 = 10
x2-4x-2x+8 3
2×2-10x+10 = 10
x2-6x+8 3
6×2-30x+30=10×2-60x+80
6×2-30x+30-10×2+60x-80=0
-4×2+30x-50=0
-2(2×2-15x+25) =0
2×2-15x+25=0
2×2-10x-5x+25=0
2x(x-5)-5(x-5)=0
(x-5) (2x-5)=0
x-5=0 2x-5=0
x=5 2x=5, x=5/2
(v) x+1 + x-2 =3
x-1 x+2
x2+2x+x+2+x2-2x-x+2 =3
(x-1)(x+2)
2×2+4 =3
x2+2x-x-2
3×2+6x-3x-6=2×2+4
3×2+3x-6-2×2-4=0
x2+3x-10=0
x2+5x-2x-10=0
x(x+5) -2(x+5)=0
(x+5)(x-2)=0
x+5=0 x-2=0
x=-5 x=2
(vi) 1 – 1 = 11
x+4 x-7 30
x-7-(x+4) = 11
(x+4)(x-7) 30
x-7-x-4 = 11
x2-7x+4x-28 30
-11 = 11
x2-3x-28 30
-1 = 1
x2-3x-28 30
x2-3x-28=-30
x2-3x-28+30=0
x2-3x+2=0
x2-2x-x+2=0
x(x-2)-1(x-2)=0
(x-2) (x-1) =0
x-2=0 x-1=0
x=2 x=1
(vii) (4x-3) -10 (2x+1) =3
2x+1 4x-3
(4x-3)2 -10(2x+1)2 = 3
(2x+1)(4x-3)
(4x)2-2•4x•3+(3)2-10(2x)2+2•2x•1+(1)2=3
8×2-6x+4x-3
16×2-24x+9-10(4×2+4x+1) =3
8×2-2x-3
16×2-24x+9-40×2-40x-10 =3
8×2-2x-3
-24×2-64x-1 =3
8×2-2x-3
-24×2-64x-1=24×2-6x-9
-24×2-64x-1-24×2+6x+9=0
-48×2-5x+8=0
24×2+29x-4=0
a=24, b=29, c=-4
D=b2-4ac=(29)2-4×24×(-4)=841+384=1225
x= -b+-√D = -29+-√1225 = -29+-35
2a 2×24 48
-29+35 = 6 = 1
48 48 8
-29 – 35 = -64 = -4
48 48 3
19. वर्ग पूरा करके हल करें——
(i) 4y2+4√3y+3=0
(2y)2+2•2y•√3+(√3)2-3+3=0
(2y+√3)2=0 , 2y+√3=(0)2=0
2y+√3=0, 2y=-√3, y=-√3/2, -√3/2
(ii) 5×2-6x-2=0
5 से गुणा करने पर,
25×2-30x-10=0
(5x)2-2•5x•3+(3)2-(3)2-10=0
(5x-3)2-9-10=0
(5x-3)2-19=0
(5x-3)2=19
5x-3=+-√19
5x=√19+3 5x= -√19+3
x= √19+3 x= -√19+3
5 5
(iii) 4y2+3y+5=0
(2y)2+2•2y 3 + (3)2 – 9 + 5
4 4 16 1
(2y+ 3 )2 -9+80
4 16
(2y+ 3 )2 + 71 = (2y+ 3 )2= -71 <0
4 16 4 16
(iv) 2z2-7z+3=0
2 से भाग देने पर,
z2- 7 z + 3 =0
2 2
z2-2•z• 7 + (7)2 – (7)2 + 3 =0
4 4 2 2
(z- 7 )2 – 49 + 3 =0
4 16 2
(z- 7 )2 – 49+48 =0
4 16
(z- 7 )2 – 1 =0
4 16
(z- 7 )2= 1
4 16
(z- 7 )= 1
4 √16
(z- 7 )= + – 1
4 4
z- 7 = 1
4 4
z= 1 + 7 = 1+7 = 8 =2
4 4 4 4
z= -1 + 7 = -1+7 = 6 = 3
4 4 4 4 2
(v) 6z2-5z-21=0
6 से गुणा करने पर,
36z2-30z-126=0
(6z)2-2•6z• 5 + (5)2 – (5)2 -126=0
2 2 2
(6z- 5 )2 – 25 – 126 =0
2 4 1
(6z- 5 ) – -25-504 =0
2 4
(6z- 5 ) = 529
2 4
(6z- 5 ) = √529
2 4
(6z- 5 ) = +- 23
2 2
6z= 23 + 5 = 23+5 = 28 =14
2 2 2 2
6z=14, z= 14 = 7
6 3
6z= -23 + 5 = -23+5 = -18 =-9
2 2 2 2
6z=-9, z= -9 = -3
6 2
(vi) 2z2+az-a2=0
2 से गुणा करने पर,
4z2+2az -2a2=0
(2z)2+2•2z• a + (a)2 – (a)2 -2a2=0
2 2 2
(2z+ a )2- a2 – 2a2 =0
2 4 1
(2z+ a )2 – a2-8a2 =0
2 4
(2z+ a )2 – 9a2 =0
2 4
(2z+ a )2= 9a2
2 4
(2z+ a )= √9a2
2 4
(2z+ a )= +- 3a
2 2
2z+ a = 3a ,= 3a – a = 2a =a
2 2 2 2 2
2z=a, z=a/2
2z + a = – 3a =- 3a – a =-4a =-2a
2 2 2 2 2
2z=-2a, z=-2a/2=-a
(vii) 12y2-25y+12=0
3 से भाग देने पर,
4y2- 25y + 4 =0
3
(2y)2-2•2y• 25 + (25)2 – (25)2 +4=0
12 12 12
(2y- 25 )2 – 625 + 4 =0
12 144 1
(2y- 25 )2 – 625+576 =0
12 144
(2y- 25 )2 – 49 =0
12 144
(2y- 25 ) = 49
12 144
2y- 25 = √49
12 144
2y- 25 = + – 7
12 12
2y- 25 = 7 = 7 + 25 = 32 = 8
12 12 12 12 12 3
2y= 8 , y= 8 = 4
3 3×2 3
2y- 25 = -7 = -7 + 25 = 18 = 3
12 12 12 12 12 2
2y= 3 , y= 3 ,y = 3
2 2×2 4
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