Find the value of k for which the equation (k-1)x^2 + (k+4)x+k+7=0 has equal roots
Answers
Answer:
Hi friend!!
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;)
The value of k for which the equation (k-1)x² + (k+4)x + k+7 = 0 has equal roots is k = 2.
Given,
The quadratic equation (k-1)x² + (k+4)x + k+7 = 0
To Find,
The value of k such such the roots of the quadratic equation are equal
Solution,
Quadratic Equation:
- The polynomial equations of degree two in one variable of type f(x) = ax² + bx + c = 0 and with a, b, c, ∈ R and a ≠ 0 are known as quadratic equations.
- It is a quadratic equation in its general form, where "a" stands for the leading coefficient and "c" for the absolute term of f(x).
- A quadratic equation has utmost 2 roots.
We have been given a quadratic equation (k-1)x² + (k+4)x + k+7 = 0.
Here, a = k-1, b = k+4, c = k+7.
We are required to find the value of k, such that the roots of the quadratic equation are equal.
We know that the roots of a quadratic equation are equal when
⇒ √(b²-4ac) = 0
on squaring both sides
⇒ b²-4ac = 0
substitute the values of a, b and c:
⇒ (k+4)² - 4(k-1)(k+7) = 0
⇒ k² + 8k + 16 - 4(k² + 7k - k -7) = 0
⇒ k² + 8k + 16 - 4(k² + 6k - 7) = 0
⇒ k² + 8k + 16 - 4k² - 24k + 28 = 0
⇒ -3k² - 16k + 44 = 0
⇒ 3k² + 16k - 44 = 0
this is also a quadratic equation. its solution is given by:
Hence, roots are 2 or 7.33
The integral value is 2.
The required value of k = 2.
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