ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.1

 Exercise 5.1

Question 1

Check whether the following are quadratic equations:

(i) 3x22x+35=0
(ii)(2x + 1) (3x – 2) = 6(x + 1) (x – 2)
(iii) (x3)3+5=x3+7x21
(iv) x3x=2,x0
(v) x+2x=x2,x0
(vi) x2+1x2=3,x0
Sol :
(i) 
3x22x+35=0

It is a quadratic equation as it is power of 2.


(ii)

(2x + 1) (3x – 2) = 6(x + 1) (x – 2)

6x² – 4x + 3x – 2 = 6(x² – 2x + x – 2)

6x² – x – 2 = 6x² – 12x + 6x – 12

12x-6x-x=-12+2

5x=-10

It is not a quadratic equation.


(iii) 

(x3)3+5=x3+7x21

x33x2×3+3x×927+5=x3+7x21

9x2+27x227x2+1=0

16x2+27x21=0

16x227x+21=0

It is a quadratic equation


(iv)

x3x=2,x0

x23=2xx22x3=0

It is a quadratic equation


(v)

x+2x=x2,x0

x2+2=x3x3x22=0

It is not a quadratic equation


(vi)

x2+1x2=3,x0
x4+6=3x2
x43x2+6=0
It is not a quadratic equation

Question 2

In each of the following, determine whether the given numbers are roots of the given equations or not;

(i) x² – x + 1 = 0; 1, – 1

(ii) x² – 5x + 6 = 0; 2, – 3

(iii) 3x² – 13x – 10 = 0; 5,23

(iv) 6x² – x – 2 = 0; 12,23

Sol :

(i)

 x² – x + 1 = 0; 1, -1

Where x = 1, then

(1)² – 1 + 1 = 1 – 1 + 1 = 1 ≠ 0

∴ x = 1 does not satisfy it

and (1)2(1)+1=0
1+1+130

∴x=-1, does not satisfy it

∴x=1,-1 are not roots of the equation


(ii)

x25x+6=0;2,3

When x=2, then

(2)25×2+6

=4-10+6=10-10=0

∴x=2 is its root

Where x=-3, then

(3)25(3)+6

=9+15+6=300

∴x=-3 is not its solution

∴2 is root of the equation by -3 is not a root.


(iii)

3x213x10=0;5,23

x=5

3(5)213×510=756510

=75-75=0

∴x=5 is it root

If x=23, then

3(23)213×2310

=3×49+26310

=43+26310=30310=1010=0

x=23 is also its root.

Hence both 5,23 are its roots.


(iv)

6x2x2=0;12,23

If x=12, then

=6(12)2(12)2

=6×14+122=32+122

=422=0

x=12 is its root

If x=23, then

=6×49232=83232

=632=0

x=23 is also its root.

Hence 12,23 are both its root.


Question 3

In each of the following, determine whether the given numbers are solutions of the given equation or not:

(i) x² – 3√3x + 6 = 0; √3, – 2√3

(ii) x² – √2x – 4 = 0, x = – √2, 2√2

Sol :

(i) x² – 3√3x + 6 = 0; √3, -2√3

(a) Substituting the value of x = √3

L.H.S. =x233x+6

=(3)233×3+6=39+6=0=R.H.S

x=3 is its solution


(b) x=23

Substituting x=23

L.H.S. =x233x+6

=(23)233(23)+6

=12+18+6=36≠0

x=23 is not its solution


(ii) x22x4=0,x=2,22

(a) x=2

Substituting x=2

L.H.S.=x22x4

=(2)22(2)4

=2+2-4=0=R.H.S
x=2 is its solution


(b) x=22

Substituting x=22

L.H.S. =x22x4

=(22)22(22)4

=8-4-4=8-8=0=R.H.S

x=22 is its solution


Question 4

(i) If 12 is a solution of the equation 3x² + 2kx – 3 = 0, find the value of k.

(ii) If 23 is a solution of the equation 7x² + kx – 3 = 0, find the value of k.

Sol : 

x=12 is a solution of the

3x² + 2kx – 3 = 0,

Substituting the value of x in the given equation

3(12)2+2k(12)3=0

3×14k3=0

34k3=0

k=343=94

Hence k=94


(ii) 7x2+kx3=0,x=23

x=23 is its solution

7(23)2+k(23)3=0

7×49+23k3=0

2893+23k=0

23k=3289

23k=2728923k=19

k=19×32=16

Hence k=16


Question 5

(i) If √2 is a root of the equation kx² + √2 – 4 = 0, find the value of k.

(ii) If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.

Sol :

(i) kx² + √2 – 4 = 0, x = √2

x = √2 is its solution

k(2)2+2×24=0
2k+24=0
2k2=0
2k=2
k=22=1
k=1

(ii) x2x(a+b)+k=0,x=a
x=a is its solution
(a)2a(a+b)+k=0
a2a2ab+k=0ab+k=0
k=ab


Question 6

If 23 and – 3 are the roots of the equation px² + 7x + q = 0, find the values of p and q.

Sol :
23 and – 3 are the roots of the equation px² + 7x + q = 0

Substituting the value of x=23 and – 3 respectively, we get

p(23)2+7(23)+q=0

49p+143+q=0

⇒4 p+42+9 q=0

⇒4p+9q=-42...(i)

and p(3)2+7(3)+q=0

⇒9p-21+q=0

⇒q=21-9p...(ii)

Substituting the value of q in (i)

⇒4 p+9(21-9 p)=-42

⇒4 p+189-81 p=-42

⇒-77 p=-42-189=-231

p=23177=3

∴q=21-9×3

=21-27=-6

∴p=3, q=-6

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