Question

find the value of k such that the quadratic equation is a perfect square x² - 2k(x-7) - 12 = 0

Answer 1

Answer:

★ QUADRATICS RESOLUTION ★

☣ FOR EQUAL ROOTS , D=0 , b² = 4ac

☣ GIVEN EQUATION → x² -2kx + (7k - 12 ) = 0

☣ APPLYING THE PRECONDITION ...

☣ 4K² = 4 [ 7K - 12 ]

☣ 4K² = 28K - 48

☣ 4K² - 28K + 48 =0

☣ K² - 7K + 12 = 0

☣ K² - 4K -3K + 12 = 0

☣ K² - 3K - 4K + 12 = 0

☣ K ( K - 3 ) - 4 ( K - 3 ) = 0

☣ [K - 4] [ K - 3] = 0

☣ HENCE ...

☣ K = 4 , 3 ARE THE REQUIRED VALUES OF "K"

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Answer 2

x^2 - 2kx + 7k - 12 = 0

if both roots are real and equals, then the equation has one solution. Then the discrminant must be zero.

==> delta = b^2 - 4ac = 0

==> 4k^2 - 4(1)(7k-12) = 0

==> 4k^2 - 28k + 48 = 0

Divide by 4:

==> k^2 - 7k + 12 = 0

==> (k -4)(k-3) = 0

Then there are two solutions:

k1 = 4

k2= 3