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

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

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

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2. Solution:

   Given that:

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

   We need to find:

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

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

   [tex]\tt{\Rightarrow D = b^2-4ac}[/tex]

   [tex]\boxed{\red{\bf{\Rightarrow b^2-4ac=0}}}\bigstar[/tex]

   [tex]\bigstar\tt{(k-5)x^2+2(k-5)x+2=0}[/tex]

   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:

   [tex]\tt{\Rightarrow b^2-4ac = 0}[/tex]

   [tex]\tt{\Rightarrow (2k-10)^2-4 \times (k-5) \times 2 = 0}[/tex]

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

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

   [tex]\tt{\Rightarrow 4k^2+100-40k-8k+40 = 0}[/tex]

   [tex]\tt{\Rightarrow 4k^2-48k+140 = 0}[/tex]

   [tex]\tt{\Rightarrow 4(k^2-12k+35) = 0}[/tex]

   [tex]\tt{\Rightarrow k^2-12k+35 = \dfrac{0}{4}}[/tex]

   [tex]\boxed{\green{\bf{\Rightarrow k^2-12k+35 = 0}}}[/tex]

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

   [tex]\tt{\Rightarrow k^2-12k+35 = 0}[/tex]

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

   [tex]\tt{\Rightarrow k^2-7k-5k+35 = 0}[/tex]

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

   [tex]\tt{\Rightarrow k(k-7)-5(k-7) = 0}[/tex]

   • Now, convert the equation in factorized form

   [tex]\tt{\Rightarrow (k-5)(k-7) = 0}[/tex]

   [tex]\tt{\Rightarrow k-5=0\:or\:k-7= 0}[/tex]

   [tex]\tt{\Rightarrow k=5\:or\:k= 7}[/tex]

   • We have obtained k value as 5 and 7.

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

   [tex]\tt{\Rightarrow (k-5)x^2+2(k-5)x+2=0}[/tex]

   [tex]\tt{\Rightarrow (5-5)x^2+2(5-5)x+2=0}[/tex]

   [tex]\tt{\Rightarrow 0x^2+0x+2=0}[/tex]

   [tex]\tt{\Rightarrow 0+0+2=0}[/tex]

   [tex]\boxed{\pink{\bf{\Rightarrow 2=0}}}[/tex]

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

   Hence, k = 7 is our required value.

   [tex]\boxed{\orange{\bf{\Rightarrow k=7}}}\checkmark[/tex]

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