(x²-7x+9)²-5(x²-7x+9)+4=0 find the value of x?
Answers
Relations between Roots and Coefficients
Formulae
Sum and product of roots
Sum of roots = + = - b/a = - (coefficient of x)/(coefficient of x²)
Product of roots = = c/a = (constant term)/(coefficient of x²)
Symmetric functions of roots
From + = -b/a and = c/a, values of other functions of and can be calculated:
( - )² = ( + )² -4 ,
² + ² = ( + )² -2 ,
³ + ³ = ( + )³ -3 ( + ),
4 + 4 = [( + )² -2 ]² -2 ( )²,
² - ² = ( + )( - ) etc.
Formation of a quadratic equation with given roots
x² -S x +P = 0, where S = sum of roots and P = product of roots.
Illustrative Examples
Example
If , are roots of the equation x² -4 x +2 = 0, find the values of
² + ²
² - ²
³ + ³
1/ + 1/
Solution
As , are roots of the equation x² -4 x +2 = 0,
+ = -(-4)/1 = 4, = 2/1 = 2
² +² = ( + )² -2 = (4)² -2 (2) = 12
² -² = ( + )( -).
Now ( -)² = ( +)² -4 = (4)² -4 (2) = 8
- = ± 8 = ±2 2
² -² = ( +)( -) = (4)(±2) = ±8
³ +³ = ( +)³ -3 ( +)
= (4)³ -3 (2)(4) = 64 -24
= 40
1/ + 1/ = ( + )/ = 4/2 =2
Example
If , are roots of the quadratic equation ax² +bx +c = 0, form an equation whose roots are
- , -
1/, 1/
Solution
Since , are roots of a x 2 +b x +c = 0,
+ = -b/a, = c/a
Here, S = (-) +(-) = -( + ) = -(-b)/a = b/a
P = (- )(-) = = c/a
Hence the required equation is x² -S x +P = 0
i.e. x² - (b/a) x + c/a = 0 i.e. a x² -b x +c = 0
Here, S = 1/ + 1/ = ( +)/ = - (b / a)/(c / a) = - b/c
P = (1/).(1/) = 1/( ) = 1/(c/a) = a/c
Hence the required equation is x² -S x +P = 0
i.e. x² -(- b/c) x + a/c = 0 i.e. cx² +bx +a = 0
Example
Find p, q if p and q are roots of the equation x² +px +q = 0.
Solution
Since p, q are roots of x² +px +q = 0, p +q = -p and pq = q.
Now pq = q => pq -q = 0 => q(p -1) = 0 => q = 0 or p = 1.
When q = 0 then p +q = -p => 2p = -q = 0 => p = 0.
When p = 1, then p +q = -p => q = -2 p = -2.
Hence the required solutions are p = q = 0 or p = 1, q = -2.
Example
If the roots of the equation 2 x² +(k +1) x +(k² -5 k +6) = 0 are of opposite signs then show that 2 < k < 3.
Solution
Since roots are of opposite signs, roots are real and distinct,
discriminant> 0 and product of roots < 0
(k +1)² -4. 2. (k² -5 k +6) > 0 and (k² -5 k +6)/2 < 0
First condition is always true when second holds. (because (k +1)² = 0)
Hence (k² -5 k +6)/2 < 0 => k² -5 k +6 < 0 => (k -2)(k -3) < 0
=> 2 < k < 3