Question

find the value of the k for which equation (k-5)x2+2(k-5)x+2=0 has real and equal roots​

Answer 1

Answer:

Step-by-step explanation:

The quadratic equation (k-5)x2 + 2(k-5)x + 2 = 0 have equal roots.

⇒ Discriminant (b2 - 4ac) = 0

⇒ [2(k-5)]2 - 4(k-5)(2) = 0

⇒ 4(k2 - 10k + 25) -(8k - 40) = 0

⇒ 4k2 - 40k + 100 - 8k + 40 = 0

⇒ 4k2 - 48k + 140 = 0

⇒ k2 - 12k + 35 = 0

⇒ (k - 7)(k - 5) = 0

⇒ k = 7 or 5

Answer 2

Solution:

Given that:

$\bigstar\tt{(k-5)x^2+2(k-5)x+2=0}$ has real and equal roots.

We need to find:

$\bigstar$ The value of k in $\tt{(k-5)x^2+2(k-5)x+2=0}$

It was given that the equation has real and equal roots. That means, the Discriminant of the equation is 0.

$\tt{\Rightarrow D = b^2-4ac}$

$\boxed{\red{\bf{\Rightarrow b^2-4ac=0}}}\bigstar$

$\bigstar\tt{(k-5)x^2+2(k-5)x+2=0}$

Let us see what are all the values of a, b and c here:

• b = 2(k-5) = 2k-10
• a = k-5
• c = 2

Let us apply the values of a, b and c in our discriminant formula to find k:

$\tt{\Rightarrow b^2-4ac = 0}$

$\tt{\Rightarrow (2k-10)^2-4 \times (k-5) \times 2 = 0}$

Let us use the identity, (a-b)² = a²+b²-2ab in $\sf{(2k-10)^2}$:

$\tt{\Rightarrow 4k^2+100-40k-8 \times (k-5) = 0}$

$\tt{\Rightarrow 4k^2+100-40k-8k+40 = 0}$

$\tt{\Rightarrow 4k^2-48k+140 = 0}$

$\tt{\Rightarrow 4(k^2-12k+35) = 0}$

$\tt{\Rightarrow k^2-12k+35 = \dfrac{0}{4}}$

$\boxed{\green{\bf{\Rightarrow k^2-12k+35 = 0}}}$

Now, we have formed a quadratic equation. Let us solve this equation by using factorisation method:

$\tt{\Rightarrow k^2-12k+35 = 0}$

• -12k can be splitted as (-7k) + (-5k). This is the first step and also called as splitting the middle term.

$\tt{\Rightarrow k^2-7k-5k+35 = 0}$

• Let us take k and -5 as common factors.

$\tt{\Rightarrow k(k-7)-5(k-7) = 0}$

• Now, convert the equation in factorized form

$\tt{\Rightarrow (k-5)(k-7) = 0}$

$\tt{\Rightarrow k-5=0\:or\:k-7= 0}$

$\tt{\Rightarrow k=5\:or\:k= 7}$

• We have obtained k value as 5 and 7.

Now, let us substitute the value of k=5 in $\sf{(k-5)x^2+2(k-5)x+2=0}$:

$\tt{\Rightarrow (k-5)x^2+2(k-5)x+2=0}$

$\tt{\Rightarrow (5-5)x^2+2(5-5)x+2=0}$

$\tt{\Rightarrow 0x^2+0x+2=0}$

$\tt{\Rightarrow 0+0+2=0}$

$\boxed{\pink{\bf{\Rightarrow 2=0}}}$

But we know that 2 is not equal to 0. So, k = 5 is rejected.

Hence, k = 7 is our required value.

$\boxed{\orange{\bf{\Rightarrow k=7}}}\checkmark$