Question

Q23-for what value of x^2 +2(k-
4) x+2k=0 has equal roots​

Answer 1

Answer :

The value of k is 2 and 8

Given :

The quadratic equation is :

• x² + 2(k - 4)x + 2k = 0
• The roots of the given equation are equal

To Find :

• The value of k

Concept to be used :

Discriminant : The discriminant of a quadratic equation ax² + bx + c is given by : b² - 4ac

Condition for type of roots according to discriminant :

• If discriminant , b² - 4ac ≥0 then the roots are real
• If discriminant , b² - 4ac = 0 , then the roots are equal
• If discriminant , b² - 4ac < 0 , then the roots are immaginary ( or no real root exists)

Solution :

Given , the equation x² + 2(k - 1)x + 2k has equal roots . So discriminant is :

$\sf \implies b^{2} - 4ac = 0 \\\\ \sf \implies \{ 2(k-4) \}^{2} - 4\times 1\times 2k = 0\\\\ \sf \implies 4(k-1)^{2} - 8k = 0 \\\\ \sf \implies 4(k^{2} - 8k + 16) -8k = 0 \\\\ \sf \implies 4k^{2} - 32k + 64 - 8k = 0 \\\\ \sf \implies 4k^{2} - 40k +64=0 \\\\ \sf \implies k^{2} - 10k + 16=0 \\\\ \sf \implies k^{2} - 2k - 8k +16= 0 \\\\ \sf \implies k(k-2)-8(k-2)=0 \\\\ \sf\implies (k-2)(k-8)=0$

Now we have :

$\sf k-2=0 \: \: and \: \: k - 8 =0 \\\\ \sf \implies k=2 \: \: and \: \implies k = 8$

Answer 2

$\rule{200}2$

$\huge\tt{QUESTION:}$

• For what value of x^2 +2(k-4) x+2k=0 has equal roots

$\rule{200}2$

$\huge\tt{CONCEPT~USED:}$

Discriminant : The discriminant of a quadratic equation ax² + bx + c is given by : b² - 4ac

Conditions for usage of discriminant :-

• If discriminant , b² - 4ac ≥0 then the roots are real
• If discriminant , b² - 4ac = 0 , then the roots are equal
• If discriminant , b² - 4ac < 0 , then the roots are immaginary ( or no real root exists)

$\rule{200}2$

$\huge\tt{SOLUTION:}$

↪b² - 4ac = 0

↪{2(k-1)}² - 4 × 1 × 2k = 0

↪4(k² - 8k + 16) - 8k = 0

↪4k² - 32k + 64 - 8k = 0

↪4k² - 40k + 64 = 0

↪k² - 10k + 16 = 0

↪k² - 2k - 8k + 16 = 0

↪k(k-2) - 8(k-2) = 0

↪(k-2) (k-8) = 0

So,

↪K - 2 = 0

↪K = 2

↪K - 8 = 0

↪k = 8

$\rule{200}2$