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For what value of \( p \), will the equation \( x^{2}-2 p x+7 p-12=0 \) have equal roots For what values of \( k \) will the equation \( x^{2}-(3 k-1) x+k=0 \) have equal roots? Find the value of \( k \) so that the equation \( k x^{2}+k x+1=-4 x^{2}-x \) have equal roo

Ask by Warren John. in Nepal
Nov 12,2024

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For the equation \( x^{2}-2 p x+7 p-12=0 \), \( p = 3 \) or \( p = 4 \). For the equation \( x^{2}-(3 k-1) x+k=0 \), \( k = 1 \) or \( k = \frac{1}{9} \). For the equation \( k x^{2}+k x+1=-4 x^{2}-x \), \( k = 5 \) or \( k = -3 \).

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To find the value of \( p \) for which the equation \( x^{2}-2 p x+7 p-12=0 \) has equal roots, we need to set the discriminant equal to zero: \[ D = b^2 - 4ac = (2p)^2 - 4(1)(7p - 12) = 4p^2 - 28p + 48 = 0. \] Solving this equation, we get \( p^2 - 7p + 12 = 0 \) which factors to \( (p-3)(p-4) = 0 \), so \( p = 3 \) or \( p = 4 \). For the second equation \( x^{2}-(3 k-1) x+k=0 \) to have equal roots, the discriminant must also be zero: \[ D = (3k - 1)^2 - 4(1)(k) = (3k - 1)^2 - 4k = 9k^2 - 6k + 1 - 4k = 9k^2 - 10k + 1 = 0. \] Using the quadratic formula, \( k = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18} \), yielding values \( k = 1 \) and \( k = \frac{1}{9} \). To find the value of \( k \) for the equation \( k x^{2}+k x+1=-4 x^{2}-x \), we can rearrange the equation to form \( (k + 4)x^{2} + (k + 1)x + 1 = 0 \). Setting the discriminant of this quadratic to zero, we have \[ D = (k + 1)^2 - 4(k + 4) = k^2 + 2k + 1 - 4k - 16 = k^2 - 2k - 15 = 0. \] Factoring gives \( (k - 5)(k + 3) = 0 \), yielding values \( k = 5 \) and \( k = -3 \). Thus, the values of \( p \) are \( 3 \) and \( 4 \), the values for \( k \) in the second equation are \( 1 \) and \( \frac{1}{9} \), and for the last equation \( k \) can be \( 5 \) or \( -3 \).

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