Question

. If sum of zeros and product of zeros of quadratic polynomial P(x) = x² + (k – 7)x + (k +1) is equal
then k=.
​

Answer 1

Equation: ${x}^{2} + (k-7)x + (k+1)$

For equal root discriminant must be equal to 0.

D = b² - 4ac , we know D = 0 (for equal roots)

b = (k-7) , a = 1 , c = (k+1)

D = 0

$=> \: \: \: {b}^{2} - 4ac = 0 \\ \\ = > \: \: \: {(k - 7)}^{2} - 4 \times 1 \times (k + 1) = 0\\ \\ = > \: \: \: {k}^{2} - 14k + 49 - 4k - 4 = 0 \\ \\ = > \: \: \: {k}^{2} - 18k + 45 = 0 \\ \\ = > \: \: \: {k}^{2} - 15k - 3k + 45 = 0 \\ \\ = > \: \: \: k(k - 15) - 3(k - 15) = 0 \\ \\ = > \: \: (k - 15)(k - 3) = 0$

Therefore, k = 3 or 15

putting k = 3,

${x}^{2} + (k - 7)x + (k + 1) = 0 \\ {x}^{2} - 4x + 4 = 0 \\ {x}^{2} - 2x - 2x + 4 = 0 \\ x(x - 2) - 2(x - 2) = 0 \\ (x - 2)(x - 2) = 0$

And put k = 15,

${x}^{2} + (k - 7)x + (k + 1) = 0 \\ {x}^{2} + 8x + 16 = 0 \\ {x}^{2} + 4x + 4x + 16 = 0 \\ x(x + 4) + 4(x + 4) = 0 \\ (x + 4)(x + 4)$

Hence, For equal roots we must have k = 3 or k = 15.

Hope, It helps you.