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
∴ Discriminant =49
∵D>0
∴ Roots are real and distinct
(ii) 2x2−3x+5=0
Here a=2, b=-3, c=5
∴D=b2−4ac=(−3)2−4×2×5=9-40=-31
∴ Discriminant =-31
∵D<0
∴ Roots are not real.
(iii) 7x2+8x+2=0
Here, a=7, b=8, c=2
∴D=b2−4ac=(8)2−4×7×2=64-56=8
∵Discriminant =8
∴ D>0
∴ Roots are real and distinct
(iv) 3x2+2x−1=0
Here a=3, b=2, c=-1
∴ D=b2−4ac=(2)2−4×3×(−1)=4+12=16
∴ Discriminant =16
∴ D>0
∴ Roots are real and distinct
(v) 16x2−40x+25=0
a=16, b=-40, c=25
∴D=b2−4ac=(−40)2−4×16×25=1600-1600=0
∴ Discriminant =0
∵ D=0
∴ Roots are real and equal.
(vi) 2x2+15x+30=0
Here, a=2, b=15, c=30
∴D=b2−4ac=(15)2−4×2×30=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) 3x2−2x+13=0
Sol :
(i) x² – 4x – 1 = 0
Here a = 1, b = -4, c = -1
∴D=b2−4ac=(−2)2−4×3×13=4−4=0
∵D=0
∴Roots are real and equal
(iii) 3x2−4√3x+4=0
Here a=3,b=−4√3,c=4
∴D=b2−4ac=(−4√3)2−4×3×4=48-48=0
∵D=0
∴Roots are real and equal
(iv) x2−12x+4=0
Here, a=1,b=−12,c=4
∴D=b2−4ac=(−12)−4×1×4=14−16=634
∵D=0
∴Roots are real and equal
(v) −2x2+x+1=0
Here, a=-2, b=1, c=1
D=b2−4ac=(1)2−4×(−2)×1=1+8=9
∵D=0
∴Roots are real and equal
(vi) 2√3x2−5x+√3=0
Here a=2√3,b=−5,c=√3
∴D=b2−4ac=(−5)2−4×2√3×√3=25−24=1
∵D=0
∴Roots are real and equal
Question 3
Find the nature of the roots of the following quadratic equations:
Here a=1,b=−12,c=−12
∴D=b2−4ac
=(−12)2−4×1×(−12)
=14+2=94
∵D=94>0
∴Roots are real and unequal
(ii) x2−2√3x−1=0
Here a=1,b=−2√3,c=−1
∴D=b2−4ac
=(−2√3)2−4×1×(−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
∵ The roots are equal
∵ D=0
⇒b2−4ac=0
⇒ 16-12p=0
⇒ 12p=16
⇒p=1612=43 ∴p=43
(ii) x2+(p−3)x+p=0
Here a=1, b=(p-3), c=p
∵ Equation has real and equal roots
∴b2−4ac=0
⇒(p−3)2−4(1)(p)=0
⇒(p−3)2−4p=0
⇒p2+9−6p−4p=0
⇒p2−10p+9=0
⇒p2−9p−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
∵ Roots are equal.
∴ D=0
⇒b2−4ac=0
∴ 16+20k=0
⇒20k=-16
⇒k=−1620=−45
Hence k=−45
(ii) (k−4)x2+2(k−4)x+4=0
Here a=k-4, b=2(k-4), c=4
D=b2−4ac
=[2(k−4)]2−4×(k−4)×4
=4(k2+16−8k)−16(k−4)
=4(k2−8k+16)−16(k−4)
=4[k2−8k+16−4k+16]
=4(k2−12k+32)
∵ Roots are equal
∴D=0
⇒4(k2−12k+32)=0
⇒k2−12k+32=0
⇒k2−8k−4k+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≠0
k≠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
=[2(m+1)]2−4×(3m+1)(m)
=4(m2+2m+1)−12m2−4m
=4m2+8m+4−12m2−4m
=−8m2+4m+4
∴ Roots are equal.
∴ D=0
⇒−8m2+4m+4=0
⇒2m2−m−1=0 (Dividing by 4)
⇒2m2−2m+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
⇒m=−12
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
∴D=b2−4ac
=k2−4×9×1=k2−36
∵Roots are equal
∴D=0
⇒k2−36=0
⇒(k+6)(k-6)=0
Either k+6=0, then k=-6
∴k-6=0, then k=6
(a) If k=6 then
9x2+6x+1=0
⇒(3x)2+2×3x×1+(1)2=0
⇒(3x+1)2=0
∵3x+1=0
⇒3x=-1
x=−13,−13
(b) If k=−6, then
9x2−6x+1=0
(3x)2−2×3x×1+(1)2=0
(3x−1)2=0⇒3x−1=0
3x=1
⇒x=13
x=13,13
(ii) x2−2kx+7k−12=0
Here a=1, b=-2 k, c=7, k-12
∴D=b2−4ac
=(−2k)2−4×1×(7k−12)
=4k2−4(7k−12)
=4k2−28k+48
∵Roots are equal
∴D=0
4k2−28k+48=0
k2−7k+12=0
k2−3k−4k+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=−b±√D2a=4k±√02×1=4×32=122=6
x=6,6
(b) If k=4, then
x=−b±√D2a=−(−2×4)±√02×1=+82=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
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
p(x2+x)+k=0 has equal roots ⇒px2+px+k=0
⇒7x2+7x+k=0
Here, a=7, b=7, c=k
b2−4ac=(7)2−4×7×k
=49-28 k
∵ Roots are equal
∴b2−4ac=0
⇒49-28 k=0
⇒ 28 k=49
⇒k=4928=74
∴k=74
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
∵ Roots are real
∴b2−4ac≥0
9-8p≥0
9≥8p
8p≤9
p≤98
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
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
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