Quadratic Equations Ex 4.3 Solutions
Question - 11 : - Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer - 11 : -
Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively
And area of the squares will be x2 and y2 respectively.
Given,
4x – 4y = 24
x – y = 6
x = y + 6
Also, x2 + y2 =468
⇒ (6 + y2) + y2 =468
⇒ 36 + y2 + 12y + y2 =468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y – 216 = 0
⇒ y2 + 18y – 12y –216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
As we know, the side of a square cannot benegative.
Hence, the sides of the squares are 12 m and(12 + 6) m = 18 m.
Question - 12 : - Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
Answer - 12 : -
(i) 2x2 –3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) Given,
2x2 – 3x + 5 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 2, b =-3 and c = 5
We know, Discriminant = b2 –4ac
= (– 3)2 – 4 (2) (5) = 9 – 40
= – 31
As you can see, b2 – 4ac <0
Therefore, no real root is possible for thegiven equation, 2x2 – 3x + 5 = 0.
Solution
(ii) 3x2 – 4√3x +4 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 3, b = -4√3 and c =4
We know, Discriminant = b2 –4ac
= (-4√3)2 – 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac =0,
Real roots exist for the given equation andthey are equal to each other.
Hence the roots will be –b/2a and –b/2a.
–b/2a = -(-4√3)/2×3 = 4√3/6= 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and2/√3.
Solution
(iii) 2x2 – 6x +3 = 0
Comparing the equation with ax2 + bx + c =0, we get
a = 2, b =-6, c = 3
As we know, Discriminant = b2 –4ac
= (-6)2 – 4 (2) (3)
= 36 – 24 = 12
As b2 – 4ac >0,
Therefore, there are distinct real roots existfor this equation, 2x2 – 6x + 3 = 0.
= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)
= (6±2√3 )/4
= (3±√3)/2
Therefore the roots for the given equation are(3+√3)/2 and (3-√3)/2