Selina Concise Class 10 Math Chapter 5 Quadratic Equations Exercise 5B Solutions
Quadratic Equations Exercise 5B
Q.1 without salving, comment upon the nature of roots of each of the following equation.
(i) 7x²-9x+2 = 0
Solution=
The given equation is 7x²-9x+2=0
Comparing given equation with the general form of quadratic equation ax²+bx+c=0
Here, a=7, b=9 and c=2
By using discriminant formula,
Dor ∆ = b²-4ac [∴∆-delta]
Put a=7, b=-9 AND c=2 in this formula
∆= (-9) ² -4 × 7 × (2)
= 81 – 56
∆ = +25
∴ D>0 or ∆ >0
Therefore, the roots of the equation is real and unequal.
* Note= * If ∆>0 then the roots of equation is real and unequal.
* If ∆=0 then the roots of equation is real and equal.
* If ∆<0 then the equation has no roots.
(ii) 6x²-13x+4=0
Solution=
The given equation is 6x²-13 x 4 = 0
Comparing given equation with the general form of quadratic equation ax²+bx+c = 0.
Here, a=6, b=-13, and c=4
By using discriminant formula,
∆ = b²=-4ac (read (∆0 = delta)
Put a=6, b= -13 and c = 4
In this formula ,
∆ = b²- 4ac
= (-13)² – 4 × 6 × 4
= 169 – 96
∆ = 73
∴ ∆ > 0 , the roots of the equation is real and unequal.
(iii) 25x²-10x+1=0
Solution=
The given equation is
25x²- 10x + 1=0
Comparing given equation with the general form of quadratic equation ax²+bx+c=0
Here , a = 25 , b = -10 , and c = 1
By using discriminant formula,
∆ = b² – 4ac
Put a = 25, b = -10, and c = 1
In this formula,
∆ = b²-4ac
= (-10) ² -4 ×25×1
= 100 -100
∆= 0 and the roots of the equation is real and equal.
(iv) x²+ 2√3x – 9=0
solution=
The given equation is x²+ 2√3x -9=0
comparing given equation with the general form of quadratic equation ax²+ bx + c = 0
Here, a = 1 , b = 2√3 and c = -9
By using discriminant formula,
∆ = b²- 4ac
Put a = 1, b = 2√3 and c = -9
In this formula,
∆ = (2√3)²-4 ×1 × (-9)
= (4×3) + 367
= 12+36
∆ = 49
∴∆ > 0
Therefore , the roots of the equation is real and unequal.
(v) x²- ax -b²=0
Solution=
The given equation is x²-ax-b²=0
Comparing given equation with the general form of quadratic equation ax²+bx+c=0
Here, a=1 ,, b=-a , and c=-b²
By using discriminant formula ,
∆=b²-4ac
Put a = 1, b = -a and c = -b² in this formula ,
∆= (-a)²- 4 × 1 × (-b²)
∆= a²+4b²
∴a²+4b² is always positive
∴∆>0
Therefore , the roots of the equation is real and unequal
(vi) 2x²+ 8x +9=0
Solution=
The given equation is 2x²+ 8x + 9 = 0
Comparing given equation with the general form of quadratic equation ax²+bx+9=0
Here, a = 2 , b =8 and c = 9
By using discriminant formula
∆= b²-4ac
Put a = 2, b = 8, and c = 9 in this formula,
∆ = (8) ² -4 × 2 × 9
= 64 – 72
∆ = -8
∴∆ < 0 and the equation has no roots.
(Q.2) (1) Find the value of ‘p’ if the following quadratic equations has equal roots.
(i) 4x² – (p-2) x +1 = 0
Solution=
The given equation
4x² – (p-2) x +1 = 0
Comparing given equation with the general form of quadratic equation ax²+bx+c = 0
Here, a = 4, b = – (p-2) and c = 1
We know that if △ = 0 then the roots of equation is equal. Means b² – 4ac = 0
Given: Roots are equal
= △ = (- (p – 2)² – 4 × 4 × 4 × 1 = 0
= (p – 2)² – 4 × 4 × 1 = 0
= P² – 2 × P × (2) + (2)² – 16 = 0
= P² – 4P + 4 – 16 = 0
= P² – 4P – 12 = 0
= P² – 6P + 2P – 12 = 0
= P (P – 6) + 2 (P – 6) = 0
Common ‘P’
= (P – 6) (P + 2) = 0 (-6 + 2 = 4)
Common (P – 6)
= P – 6 = 0 or P + 2 = 0
= P = 6 or P = 2
(ii) x² + (P – 3) x + P = 0
= Solution:
The given equation is x² + (P – 3) x + P = 0
Comparing given equation with the general form of quadratic equation ax² + bx + c = 0
Here, a = 1, b = (P – 3) and c = P
We know that,
If △ = 0 then the roots of equation is equal means b² – 4ac = 0
Given: the roots are equal
△ = b² – 4ac = 0
= (P – 3)² – 4 × 1 × P = 0
= P² – 2 × P × 3 + (3)² – 4P = 0
= P² – 6P + 9 – 4P = 0
= P² – 10P + 9 = 0
Using factorization method
= P² – 9P – P + 9 = 0
= P (P – 9) – 1 (P – 9) = 0
Common ‘P’
= (P – 9) (P – 1) = 0 (-9 – 1 = -10)
Common (P – 9)
= P – 9 = 0 or P – 1 = 0
= P = 9 or P = 1