5. Find the value of k, such that the number
given against each equation is one of its roots:
(i) x2 + kx - 21 = 0; 3
(ii) x2 + kx - 5 = 0; 1
(iii) x2 + 12x + k = 0; -5
Answers
Answer:
(i) K = 4.
(ii) k = 4.
(iii) k = 35.
Step-by-step explanation:
(i) x² + kx - 21 = 0 ; 3
If 3 is root of equation x² + kx - 21 = 0, Then x = 3.
Now, put x = 3 in equation :
→ (3)² + k × (3) -21 = 0
→ 9 + 3k - 21 = 0
→ 9 + 3k = 21
→ 3k = 21 - 9
→ 3k = 12
→ k = 12/3
→ k = 4
Thus,
Value of k is 4.
(ii) x² + kx - 5 = 0 ; 1
If 1 is root of equation x² + kx - 5 = 0, then x = 1.
Now, Put x = 1 in equation :
→ (1)² + k × (1) - 5 = 0
→ 1 + k - 5 = 0
→ 1 + k = 0 + 5
→ k = 5 - 1
→ k = 4
Thus,
Value of k is 4.
(iii) x² + 12x + k = 0 ; -5
If -5 is root of equation x² + 12x + k = 0 ,Then x = -5.
Now, Put x = -5 in equation :
→ (-5)² + 12 × (-5) + k = 0
→ 25 + (-60) + k = 0
→ 25 - 60 + k = 0
→ -35 + k = 0
→ k = 35
Thus,
Value of k is 35.
Answer:
(i) k=4
(ii) k=4
(iii) k=35
Step-by-step explanation:
Concept used:
If one zero is given,then substitute the value of one given zero in the given quadratic equation and then find the value of k.
FIRST PART:
f(x) =x²+kx-21=0
substitute X as 3
f(3)=3²+3k-21=0
=> f(3)=9+3k-21=0
=> 3k-12=0
=>3 k=12
=>k=12÷3
=>k=4
SECOND PART:
f(x) =x²+kx-5=0
Substitute the value of x as 1
f(1)=1+k-5=0
=>k=4=0
=>1k=4
THIRD PART:
f(x) =x²+12x+k=0
Substitute the value of x as 5
f(5)=(-5)²+12(-5)+k=0
=> 25-60+k=0
=>k-35=0
=>k=35