Quadratic Equations – Exercise 9.8 – Questions:
- Form the quadratic equation whose roots are
- i) 3, 5
- ii) 6, -5
iii) -3, 3/2
- iv) 2/3 , 3/2
- v) (2 + √3), (2 – √3)
- vi) (-3 + 2√5)(-3 – 2√5)
B.
- If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of
(i) (m + n)mn
(ii) 1/m + 1/n
(iii) m3n2 + n3m2
(iv) 1/n – 1/m
- IF a and b are the roots of the equation 3m2 = 6m + 5, find the value of
(i) a/b + b/a
(ii) (a + 2b)(2a + b)
- If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of
(i) (p + q)2 + 4pq
(ii) p3 + q3
- Form a quadratic equation whose roots are p/q and q/p
- Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root which is twice the other.
- Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.
- Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.
- If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.
Quadratic Equations – Exercise 9.8 – Solution:
- Form the quadratic equation whose roots are
i) 3, 5
Solution:
Let m and n be the roots
Then m = 3 and n = 5
Sum of the roots = m + n = 3 + 5 = 8
Product of the roots = mn = 3 x 5 = 15
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – 8x + 15 = 0
ii) 6, -5
Solution:
Let m and n be the roots
Then m = 6 and n = -5
Sum of the roots = m + n = 6 – 5 = 1
Product of the roots = mn = 6 x (-5) = -30
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – 1x + (-30) = 0
x2 – x – 30 = 0
iii) -3, 3/2
Solution:
Let m and n be the roots
Then m = -3 and n = 3/2
Sum of the roots = m + n = -3 + 3/2 = -6+3/2 = –3/2
Product of the roots = mn = -3 x 3/2 = -9/2
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – (-3/2) x + (-9/2) = 0
x2 + 3/2x – 9/2 = 0
iv) 2/3 , 3/2
Solution:
Let m and n be the roots
Then m = 2/3 and n = 3/2
Sum of the roots = m + n = 2/3 + 3/2 = 4+9/6 = 13/6
Product of the roots = mn = 2/3 x 3/2 =1
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – (13/6) x + (1) = 0
x2 – 13/6x + 1 = 0
v) (2 + √3), (2 – √3)
Solution:
Let m and n be the roots
Then m = 2 + √3 and n = 2 – √3
Sum of the roots = m + n = (2 + √3)+( 2 – √3) = 4
Product of the roots = mn = (2 + √3)( 2 – √3) = 4 – 2√3 + 2√3 – 3 = 4 – 3 = 1
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – (4) x + (1) = 0
x2 – 4x + 1 = 0
vi) (-3 + 2√5)(-3 – 2√5)
Solution:
Let m and n be the roots
Then m = -3 + 2√5 and n = -3 – 2√5
Sum of the roots = m + n = (-3 + 2√5)+( -3 – 2√5) = -6
Product of the roots = mn = (-3 + 2√5)( -3 – 2√5) = 9 + 6√5 – 6√5 – 4×5 = 9 – 45 = – 36
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – (-6) x + (-36) = 0
x2 + 6x – 36 = 0
B.
- If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of
(i) (m + n)mn
(ii) 1/m + 1/n
(iii) m3n2 + n3m2
(iv) 1/n – 1/m
Solution:
The given quadratic equation is x2 – 6x + 2 = 0 which is of the form ax2 + bx + c = 0 where a = 1, b = -6 and c = 2
Therefore, m = 3 + √7 and n = 3 – √7
(i) (m + n)mn
= (3+ √7+ 3 – √7)( 3 + √7)( 3 – √7)
= 6 (3 + √7)( 3 – √7)
=6 (9 – 3√7 + 3√7 – 7)
= 6 x 2
= 12
(ii) 1/m + 1/n
= 1/(3 + √7) + 1/(3 – √7)
= (3 – √7)+(3 + √7)/(3 – √7)(3+√7)
= 6/2
= 3
(iii) m3n2 + n3m2
= (3 + √7)3( 3 – √7)2 + (3 – √7)3( 3 + √7)2
= (90 + 34√7)(16 – 6√7) + (90 – 34√7)(16 + 6√7)
= 12 + 4√7 + 12 – 4√7
= 24
(iv) 1/n – 1/m
= 1/(3 – √7) – 1/(3 + √7)
= (3 + √7)-(3 – √7)/(3 – √7)(3+√7)
= 2√7/2
= √7
- If a and b are the roots of the equation 3m2 = 6m + 5, find the value of
(i) a/b + b/a
(ii) (a + 2b)(2a + b)
Solution:
The given equation is 3m2 = 6m + 5, this can be written as 3m2 – 6m – 5 = 0, which is of the form mx2 + nx + p = 0 where m = 3 ; n = -6 and p = -5
- If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of
(i) (p + q)2 + 4pq
(ii) p3 + q3
Solution:
Given p and q are the roots of the equation is 2a2 – 4a + 1 = 0, which is of the form ax2 + bx + c = 0 where a = 2, b = -4 , x = 1
- Form a quadratic equation whose roots are p/q and q/p
Solution:
Let p/q and q/p be the roots of a quadratic equation
Sum of the roots = m + n = p/q + q/p = p^2+q^2/pq
Product of the roots = mn = p/q x q/p = 1
Standard form is x2 –(m+n)x + mn = 0
Thus, x2 – (p^2+q^2/pq) x + (1) = 0
pqx2 – (p2 + q2)x – pq = 0
- Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root equal to zero.
Solution:
The given quadratic equation is x2 + 4x + (k + 2) = 0 where a = 1, b = 4 , c = k+2
The sum of the roots m+n = –b/a = – 4/1 = -4
The product of the roots mn = c/a = k+2/1 = k + 2
If one root is m then other root is zero.
Thus, m = m and n = 0
Therefore, m + n = – 4 ⇒ m + 0 = -4 ⇒m = -4
mn = k + 2 ⇒ 0 = k + 2 ⇒ k = -2
- Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.
Solution:
The given quadratic equation is 2x2 – 3qx + 5q = 0 where a = 2, b = -3q , c = 5q
The sum of the roots m+n = –(-3q)/2 = 3q/2
The product of the roots mn = c/a = 5q/2
If one root is m then other root is 2m.
Thus, m = m and n = 2m
Therefore, m + 2m = 3q/2 ⇒ 3m = 3q/2 ⇒m = q/2
mn = 5q/2 ⇒ 2m2 = (5q/2)
⇒ 2(q/2)2 = 5q/2
⇒ 2(q^2/4) = 5q/2
⇒ q^2/4 = 5q/4
⇒q2 = 5q
⇒q = 5
- Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.
Solution:
The given quadratic equation is 4x2 – 8px + 9 = 0 where a = 4, b = -8p , c = 9
The sum of the roots m+n = –(-8p)/4 = 2p
The product of the roots mn = c/a = 9/4
If one root is m then other root is (m – 4).
Thus, m = m and n = (m – 4)
m+n = 2p⇒ m+m-4 = 2p⇒ 2m = 2p + 4⇒m = p + 2
mn = 9/4 ⇒m(m – 4) = 9/4
⇒m2 – 4m – 9/4 = 0
⇒(p+2)2 – 4(p+2) – 9/4 = 0
⇒p2 + 4p + 4 – 4p – 8 – 9/4 = 0
⇒p2 –25/4 = 0
⇒p2 = 25/4
⇒p = ±5/2
- If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.
Solution:
The given quadratic equation is x2 + px + q = 0 where a = 1, b = p , c = q
The sum of the roots m+n = –(p)/1 = -p
The product of the roots mn = q/1 = q
If one root is m then other root is 3m.
Thus, m = m and n = 3m
m + n = -p ⇒ m+3m = -p
⇒4m = -p
⇒ m = -p/4
mn = q ⇒m(3m) = q
⇒3m2 = q
⇒3(-p/4)2 = q
⇒3p^2/16 = q
⇒3p2 = 16q