solve : (k+4)x2+(k+1)x +1=0 has equal roots . find the value of k.also find the roots
Answers
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
I hope this will help you;)
Discriminant = b^2 - 4ac
Discriminant = (k+1)^2 - 4(1)(k+4)
= (k+1)^2 - 4(1)(k+4) = k^2 +1+2k -4k - 16 = k^2 - 2k - 15
For equal roots, discriminant = 0
→ k^2 - 2k - 15 = 0
→ k^2 - 5k + 3k - 15 = 0
→ k(k - 5) + 3(k - 5) = 0
→ (k - 5)(k + 3) = 0
→ k = 5 or k = -3
k = 5 or k = - 3
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Taking , k = 5
→ (k + 4)x^2 + (k + 1)x + 1 = 0
→ (5 + 4)x^2 + (5 + 1)x + 1 = 0
→ 9x^2 + 6x + 1 = 0
→ (3x + 1)^2 = 0
x = -1/3 [root]
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Taking x = -3
→ (-3+ 4)x^2 + (-3 + 1)x + 1 = 0
→ (1)x^2 + (-2)x + 1 =0
→ x^2 - 2x + 1 = 0
→ (x -1)^2 = 0
→ x= 1
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Roots are 1 or -1/3 .