Selina Concise Class 10 Math Chapter 5 Quadratic Equations Exercise 5A Solutions
Quadratic Equations
(Q1) Find which of the following are quadratic”
(1) (3x – 1)² = 5 (x + 8)
=> Solution:
Given equation is
(3x – 1)² = 5 (x + 8)
= > (3x)² – 2×(3x) × (-1) + (1)² = 5x + 40
= > 9x² – [-6x] + 1 = 5x + 40
[∵ (a-b)² = a² – 2ab + b²]
=> 9x² + 6x + 1 = 5x + 40
=> 9x² + 6x – 5x + 1 – 40 = 0
=> 9x² + x – 39 = 0
∴ The general form of quadratic equation is ax² + bx + c = 0
x is a variable and a, b, c are real numbers.
∴ The solution is in the general form of quadratic equation.
∴ The given equation is a quadratic equation.
(2) 5x² – 8x = -3 (7 – 2x)
= Solution:
The given equation is
5x² – 8x = -3 (7 – 2x)
= >5x² – 8x = -3×7 + 3×2x
= > 5x² – 8x = – 21 + 6x
= > 5x² – 8x – 6x + 21 = 0
= > 5x² – 14x + 21 = 0
∴ The general form of quadratic equation is ax² + bx + c = 0
x is a variable and a, b, and c are real numbers.
∴ The solution is in the general form of quadratic equation.
∴ The given equation is a quadratic equation.
(3) (x – 4) (3x + 1) = (3x – 1) (x + 2)
= Solution:
The given equation is (x – 4) (3x + 1) = (3x – 1) (x + 2)
Multiply two brackets.
= > x × (3x + 1) – 4 (3x + 1) = 3x × (x + 2) – 1 (x + 2)
= > 3x² + x – 12x – 4 = 3x² + 6x – x – 2
On both sides ‘3x²’ term is same, so 3x² get cancel on both sides.
=> x – 12x – 4 = 6x – x – 2
=> x – 6x – 12x + x – 4 + 2 = 0
=> -5x – 12x + x – 2 = 0
=> – 16x – 2 = 0
∴ The general form of quadratic equation is ax² + bx + c = 0
∴ x is a variable but b and c are present in solution.
∴ The solution is not in the general form of quadratic equation.
∴ The given equation is not a quadratic equation.
(4) x² + 5x – 5 = (x – 3)²
= Solution:
The given equation is
x² + 5x – 5 = (x – 3)²
= > x² + 5x – 5 = x² – 2 × x × (-3) + (-3)²
= > x² + 5x – 5 = x² + 6x + 9 [∵ (a – b)² = a² – 2ab + b²]
= On both sides ‘x²’ term is same. So x² get cancel on both sides.
= 5x – 5 = 6x + 9
= 5x – 6x – 5 – 9 = 0
= -x – 14 = 0
∴ The general form of quadratic equation is ax² + bx + c = 0
∴ x is a variable but second & third term are present and first is not present in solution.
∴ Means bx + c are present ax² is not present
∴ The given equation is not a quadratic equation.
∴ In quadratic equation the degree is ‘2’
(5) 7x³ – 2x² + 10 = (2x – 5)²
= Solution:
The given equation is
7x³ – 2x² + 10 = (2x – 5)²
=> 7x³ – 2x² + 10 = (2x)² – 2 (2x) × (-5) + (-5)²
=> 7x³ – 2x² + 10 = 4x² + 20x + 25
=> 7x³ – 2x² – 4x² – 20x + 10 – 25 = 0
=> 7x³ – 6x² – 20x – 15 = 0
∴ The general form of quadratic equation is ax² + bx + c = 0
∴ The quadratic equation degree is ‘2’ but the given equation the degree is ‘3’ and the equation is cubic equation.
∴ The given equation is not a quadratic equation.
(6) (x – 2)² + (x + 2)² + 3(x + 1) = 0
= Solution:
The given equation is
(x – 1)² + (x + 2)² + 3 (x + 1) = 0
=> x² – 2 × x ×1-(1) ²+ x²+2 × x × 2 + (2) ²+3x +3 = 0
=> x² -2x -1 + x² + 4x + 4 + 3x +3=0
=> 2x²+2x- 1 +4+ 3x+3 = 0
=> 2x²+5x+3+3=0
=> 2x²+5x+6=0
∴ The general form of quadratic education is 2x²+5x+6=0.
∴x is a variable and a, b and c are real number.
∴ The solution is in the general form of quadratic equation.
∴ The given equation is quadratic equation.
Q.2
(1) IS x=5 a solution of the quadratic equation x²-2x-15=0?
Solution=
The given equation is
x² – 2x- 15 = 0
∴ x = 5 substitute in given equation
L.H.S = (5) ²-2×5-15
= 25-10-15
=25-25
= 0
= R.H.S
∴ L.H.S = R.H.S
∴ x=5 is a solution of given equation x²-2x-15= 0
(2) Is x = -3 a solution of the quadratic equation 2x²-7x=9=0?
= solution =
The given quadratic equation is 2x²-7x+9 = 0
Substitute x=-3 in given equation
L.H.S =2x²-7x + 9
= 2× (-3)² -7 (-3)+9
=2 × (9) + 21+ 9
= 18 + 21 + 9
= 48
≠ R.H.S
∴ L.H.S ≠ R.H.S
∴x=-3 is not solution of given quadratic equation 2x²-7x+9=0