ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.4

 Exercise 5.4

Question 1

Find the discriminant of the following equations and hence find the nature of roots:

(i) 3x² – 5x – 2 = 0

(ii) 2x² – 3x + 5 = 0

(iii) 7x² + 8x + 2 = 0

(iv) 3x² + 2x – 1 = 0

(v) 16x² – 40x + 25 = 0

(vi) 2x² + 15x + 30 = 0.

Sol :

(i) 3x² – 5x – 2 = 0

Here a = 3, b = -5, c = -2

$\therefore \mathrm{D}=b^{2}-4 a c=(-5)^{2}-4 \times 3 \times(-2)$=25+24=49

∴ Discriminant =49

∵D>0

∴ Roots are real and distinct


(ii) $2 x^{2}-3 x+5=0 $

Here a=2, b=-3, c=5

$\therefore  D=b^{2}-4 a c=(-3)^{2}-4 \times 2 \times 5$=9-40=-31

∴ Discriminant =-31

∵D<0

∴ Roots are not real.


(iii) $7 x^{2}+8 x+2=0$

Here, a=7, b=8, c=2

$\therefore D=b^{2}-4 a c=(8)^{2}-4 \times 7 \times 2$=64-56=8

∵Discriminant $=8$

∴ D>0

∴ Roots are real and distinct


(iv) $3 x^{2}+2 x-1=0$

Here a=3, b=2, c=-1

∴ $D=b^{2}-4 a c=(2)^{2}-4 \times 3 \times(-1)$=4+12=16

∴  Discriminant =16

∴ D>0

∴  Roots are real and distinct


(v) $16 x^{2}-40 x+25=0$

a=16, b=-40, c=25

$\therefore \mathrm{D}=b^{2}-4 a c=(-40)^{2}-4 \times 16 \times 25$=1600-1600=0

∴  Discriminant =0

∵ D=0

∴  Roots are real and equal.


(vi) $2 x^{2}+15 x+30=0$

Here, a=2, b=15, c=30

$\therefore D=b^{2}-4 a c=(15)^{2}-4 \times 2 \times 30$=225-240=-15

∴ Discriminant =-15

∵ D<0

∴ Root are not real.


Question 2

Discuss the nature of the roots of the following quadratic equations :

(i) x² – 4x – 1 = 0

(ii) $3 x^{2}-2 x+\frac{1}{3}=0$

(iii) $3 x^{2}-4 \sqrt{3 x}+4=0$

(iv) $x^{2}-\frac{1}{2} x+4=0$

(v) $-2 x^{2}+x+1=0$

(vi) $2 \sqrt{3} x^{2}-5 x+\sqrt{3}=0$

Sol :

(i) x² – 4x – 1 = 0

Here a = 1, b = -4, c = -1

$\therefore D=b^{2}-4 a c=(-4)^{2}-4 \times 1 \times(-1)$=16+4=20
∵D>0
Roots are real and distinct

(ii) $3 x^{2}-2 x+\frac{1}{3}=0$
Here a=3, b=-2, $c=\frac{1}{3}$

$\therefore D=b^{2}-4 a c=(-2)^{2}-4 \times 3 \times \frac{1}{3}=4-4=0$

∵D=0

∴Roots are real and equal


(iii) $3 x^{2}-4 \sqrt{3} x+4=0$

Here $a=3, b=-4 \sqrt{3}, c=4$

$\therefore D=b^{2}-4 a c=(-4 \sqrt{3})^{2}-4 \times 3 \times 4$=48-48=0

∵D=0

∴Roots are real and equal


(iv) $x^{2}-\frac{1}{2} x+4=0$

Here, $a=1, b=-\frac{1}{2}, c=4$

$\therefore \mathrm{D}=b^{2}-4 a c=\left(-\frac{1}{2}\right)-4 \times 1 \times 4=\frac{1}{4}-16=\frac{63}{4}$

∵D=0

∴Roots are real and equal


(v) $-2 x^{2}+x+1=0$

Here, a=-2, b=1, c=1

$\mathrm{D}=b^{2}-4 a c=(1)^{2}-4 \times(-2) \times 1$=1+8=9

∵D=0

∴Roots are real and equal


(vi) $2 \sqrt{3} x^{2}-5 x+\sqrt{3}=0$

Here $a=2 \sqrt{3}, b=-5, c=\sqrt{3}$

$\therefore \mathrm{D}=b^{2}-4 a c=(-5)^{2}-4 \times 2 \sqrt{3} \times \sqrt{3}=25-24=1$

∵D=0

∴Roots are real and equal


Question 3

Find the nature of the roots of the following quadratic equations:

(i) $x^{2}-\frac{1}{2} x-\frac{1}{2}=0$
(ii) x² – 2√3x – 1 = 0 If real roots exist, find them.
Sol :
(i) $x^{2}-\frac{1}{2} x-\frac{1}{2}=0$

Here $a=1, b=-\frac{1}{2}, c=-\frac{1}{2}$

$\therefore \mathrm{D}=b^{2}-4 a c$

$=\left(\frac{-1}{2}\right)^{2}-4 \times 1 \times\left(\frac{-1}{2}\right)$

$=\frac{1}{4}+2=\frac{9}{4}$

$\because D=\frac{9}{4}>0$

∴Roots are real and unequal


(ii) $x^{2}-2 \sqrt{3} x-1=0$

Here $a=1, b=-2 \sqrt{3}, c=-1$

$\therefore D=b^{2}-4 a c$

$=(-2 \sqrt{3})^{2}-4 \times 1 \times(-1)=12+4=16$

∵ D>0

∴ Roots are real and unequal.


Question 4

Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots:

(i) px² – 4x + 3 = 0

(ii) x² + (p – 2)x + p = 0.

Sol :

(i) px² – 4x + 3 = 0

Here a = p, b = -4, c = 3

$\therefore \mathrm{D}=b^{2}-4 a c=(-4)^{2}-4 \times p \times 3$=16-12 p

∵ The roots are equal 

∵ D=0

$\Rightarrow b^{2}-4 a c=0$

⇒ 16-12p=0

⇒ 12p=16

$\Rightarrow p=\frac{16}{12}=\frac{4}{3}$ $ \therefore \quad p=\frac{4}{3}$


(ii) $x^{2}+(p-3) x+p=0$

Here a=1, b=(p-3), c=p

∵ Equation has real and equal roots 

$\therefore b^{2}-4 a c=0$

$\Rightarrow(p-3)^{2}-4(1)(p)=0 $

$\Rightarrow(p-3)^{2}-4 p=0$

$\Rightarrow p^{2}+9-6 p-4 p=0 $

$\Rightarrow p^{2}-10 p+9=0$

$\Rightarrow p^{2}-9 p-p+9=0 $

⇒p(p-9)-1(p-9)=0

⇒(p-1)(p-9)=0

∴ p=1,9


Question 5

Find the value (s) of k for which each of the following quadratic equation has equal roots :

(i) kx² – 4x – 5 = 0

(ii) (k – 4) x² + 2(k – 4) x + 4 = 0

Sol :

(i) kx² – 4x – 5 = 0

Here a = k, b = -4, c = 5

∴$\mathrm{D}=b^{2}-4 a c=(-4)^{2}-4 \times k \times(-5)$=16+20 k

∵ Roots are equal. 

∴ D=0

⇒$b^{2}-4 a c=0$

∴ 16+20k=0 

⇒20k=-16

$\Rightarrow k=\frac{-16}{20}=\frac{-4}{5}$

Hence $k=\frac{-4}{5}$


(ii) $(k-4) x^{2}+2(k-4) x+4=0$

Here a=k-4, b=2(k-4), c=4

$\mathrm{D}=b^{2}-4 a c$

$=[2(k-4)]^{2}-4 \times(k-4) \times 4$

$=4\left(k^{2}+16-8 k\right)-16(k-4)$

$=4\left(k^{2}-8 k+16\right)-16(k-4)$

$=4\left[k^{2}-8 k+16-4 k+16\right]$

$=4\left(k^{2}-12 k+32\right)$

∵ Roots are equal 

∴D=0

$\Rightarrow 4\left(k^{2}-12 k+32\right)=0$

$\Rightarrow k^{2}-12 k+32=0$

$\Rightarrow k^{2}-8 k-4 k+32=0$

⇒k(k-8)-4(k-8)=0

⇒(k-8)(k-4)=0

Either k-8=0, then k=8

or k-4=0 then k=4

But $k-4 \neq 0$

$k \neq 4$

k=8


Question 6

Find the value(s) of m for which each of the following quadratic equation has real and equal roots:

(i) (3m + 1)x² + 2(m + 1)x + m = 0

(ii) x² + 2(m – 1) x + (m + 5) = 0

Sol :

(i) (3m + 1)x² + 2(m + 1)x + m = 0

Here a = 3m + 1, b = 2(m + 1), c = m

$\therefore \mathrm{D}=b^{2}-4 a c$

$=[2(m+1)]^{2}-4 \times(3 m+1)(m)$

$=4\left(m^{2}+2 m+1\right)-12 m^{2}-4 m$

$=4 m^{2}+8 m+4-12 m^{2}-4 m$

$=-8 m^{2}+4 m+4$

∴ Roots are equal. 

∴ D=0

$\Rightarrow-8 m^{2}+4 m+4=0$

$\Rightarrow 2 m^{2}-m-1=0$ (Dividing by 4)

$\Rightarrow 2 m^{2}-2 m+m-1=0$

⇒ 2m(m-1)+1(m-1)=0

⇒(m-1)(2 m+1)=0

Either m-1=0, then m=1 or 2m+1=0, then 2 m=-1

$\Rightarrow m=-\frac{1}{2}$


Question 7

Find the values of k for which each of the following quadratic equation has equal roots:

(i) 9x² + kx + 1 = 0

(ii) x² – 2kx + 7k – 12 = 0

Also, find the roots for those values of k in each case.

Sol :

(i) 9x² + kx + 1 = 0

Here a = 9, b = k, c = 1

$\therefore \mathrm{D}=b^{2}-4 a c$

$=k^{2}-4 \times 9 \times 1=k^{2}-36$

∵Roots are equal

∴D=0

$\Rightarrow k^{2}-36=0 $

⇒(k+6)(k-6)=0

Either k+6=0, then k=-6

∴k-6=0, then k=6

(a) If k=6 then

$9 x^{2}+6 x+1=0$

$\Rightarrow(3 x)^{2}+2 \times 3 x \times 1+(1)^{2}=0$

$\Rightarrow(3 x+1)^{2}=0$

∵3x+1=0

⇒3x=-1

$x=-\frac{1}{3},-\frac{1}{3}$


(b) If $k=-6,$ then

$9 x^{2}-6 x+1=0$

$(3 x)^{2}-2 \times 3 x \times 1+(1)^{2}=0$

$(3 x-1)^{2}=0 \Rightarrow 3 x-1=0$

3x=1 

$\Rightarrow x=\frac{1}{3}$

$x=\frac{1}{3}, \frac{1}{3}$


(ii) $x^{2}-2 k x+7 k-12=0$

Here a=1, b=-2 k, c=7, k-12

∴$\mathrm{D}=b^{2}-4 a c$

$=(-2 k)^{2}-4 \times 1 \times(7 k-12)$

$=4 k^{2}-4(7 k-12)$

$=4 k^{2}-28 k+48$

∵Roots are equal 

∴D=0

$4 k^{2}-28 k+48=0$

$k^{2}-7 k+12=0$

$k^{2}-3 k-4 k+12=0$

k(k-3)-4(k-3)=0

(k-3)(k-4)=0

Either k-3=0, then  k=3

or k-4=0, then k=4


(a) If k=3, then

$x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{4 k \pm \sqrt{0}}{2 \times 1}=\frac{4 \times 3}{2}=\frac{12}{2}=6$

x=6,6

(b) If k=4, then

$x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-2 \times 4) \pm \sqrt{0}}{2 \times 1}=\frac{+8}{2}=4$

∴x=4,4


Question 8

Find the value(s) of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.

Sol :

The quadratic equation given is (2p+1)x²–(7p+2)x+(7p–3)=0

Comparing with ax² + bx + c = 0, we have

a=2 p+1, b=-(7 p+2), c=(7 p-3)
$\mathrm{D}=b^{2}-4 a c$
$ \Rightarrow 0=[-(7 p+2)]^{2}-4(2 p+1)$
(7 p-3)
$0=49 p^{2}+4+28 p-4\left(14 p^{2}-6 p+7 p-3\right)$
$0=49 p^{2}+4+28 p-56 p^{2}-4 p+12$
$0=-7 p^{2}+24 p+16$
$0=-7 p^{2}+28 p-4 p+16$
0=-7 p(p-4)-4(p-4)
0=(-7 p-4)(p-4)
-7 p-4=0 or p-4=0
Hence, the value of $p=\frac{-4}{7}$ or p=4


Question 9

If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Sol :

-5 is a root of the quadratic equation

2x² + px – 15 = 0, then

⇒ 2(5)² – p( -5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 35 – 5p = 0

⇒ 5 p=35
$ \Rightarrow p=\frac{35}{5}=7$

$p\left(x^{2}+x\right)+k=0$ has equal roots $\Rightarrow p x^{2}+p x+k=0$

$\Rightarrow 7 x^{2}+7 x+k=0$

Here, a=7, b=7, c=k

$b^{2}-4 a c=(7)^{2}-4 \times 7 \times k$

=49-28 k

∵ Roots are equal 

$\therefore b^{2}-4 a c=0$

⇒49-28 k=0

⇒ 28 k=49 

$\Rightarrow k=\frac{49}{28}=\frac{7}{4}$

$\therefore k=\frac{7}{4}$


Question 10

Find the value(s) of p for which the equation 2x² + 3x + p = 0 has real roots.

Sol :

2x² + 3x + p = 0

Here, a = 2, b = 3, c = p

$b^{2}-4 a c=(3)^{2}-4 \times 2^{\circ} \times p$
=9-8 p

∵ Roots are real 

∴$b^{2}-4 a c \geq 0$

9-8p≥0

9≥8p

8p≤9

$p \leq \frac{9}{8}$


Question 11

Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.

Sol :

x² + kx + 4 = 0

Here, a = 1, b = k, c = 4

$b^{2}-4 a c=k^{2}-4 \times 1 \times 4$
$=k^{2}-16$
∵Roots are real and positive
∴$k^{2}-16 \geq 0 \Rightarrow k^{2} \geq 16$
k≥4
k=4

Question 12

Find the values of p for which the equation 3x² – px + 5 = 0 has real roots.

Sol :

3x² – px + 5 = 0

Here, a = 3, b = -p, c = 5

$\therefore b^{2}-4 a c=(-p)^{2}-4 \times 3 \times 5$
$=p^{2}-60$
∴Roots are real
∴$b^{2}-4 a c \geq 0$
∴$p^{2}-60 \geq 0 \Rightarrow p^{2} \geq 60$
$\Rightarrow p \geq \pm \sqrt{60}=\pm 2 \sqrt{15}$
$\therefore p \leq-2 \sqrt{15}$ or $p \geq 2 \sqrt{15}$

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