solve using factorization.
(k + 4)x² + (k + 1)x + 1 = 0
Answers
Answer:
We know that, for a given equation ax²+bx+c=0 to have equal roots, the required condition is
b²-4ac=0
Given,
(k+4)x²+(k+1)x +1=0 has equal roots
→(k+1)²-4(k+4)=0
k²+1+2k-4k-16=0
k²-2k-15=0
k²-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=0
k=-3,5
If k=-3
Now, the polynomial becomes
(k+4)x²+(k+1)x +1=0
(-3+4)x²+(-3+1)x+1=0
x²-2x+1=0
(x-1)²=0
x=1
If k=5
Now, the polynomial becomes
(k+4)x²+(k+1)x +1=0
9x²+6x+1=0
9x²+3x+3x+1=0
3x(3x+1)+1(3x+1)=0
(3x+1)²=0
3x+1=0
x=-1/3
Now, the roots are 1 (or)-1/3
Answer:
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Step-by-step explanation:
Answer:
We know that, for a given equation ax²+bx+c=0 to have equal roots, the required condition is
b²-4ac=0
Given,
(k+4)x²+(k+1)x +1=0 has equal roots
→(k+1)²-4(k+4)=0
k²+1+2k-4k-16=0
k²-2k-15=0
k²-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=0
k=-3,5
If k=-3
Now, the polynomial becomes
(k+4)x²+(k+1)x +1=0
(-3+4)x²+(-3+1)x+1=0
x²-2x+1=0
(x-1)²=0
x=1
If k=5
Now, the polynomial becomes
(k+4)x²+(k+1)x +1=0
9x²+6x+1=0
9x²+3x+3x+1=0
3x(3x+1)+1(3x+1)=0
(3x+1)²=0
3x+1=0
x=-1/3
Now, the roots are 1 (or)-1/3