Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes throug
the mid-point of the line segm

1 thought on “<br />Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes throug<br />the mid-point of the line segm”

1. Given :-

   If α and β are the zeros of the quadratic polynomial f(x) = kx² + 4x + 4 such that α² + β² = 24,

   To Find :-

   Value of k

   Solution :-

   Sum of zeroes

   \pmb { \alpha + \beta = \dfrac{ – b}{a}}

   α+β=

   a

   −b

   α+β=

   a

   −b

   \alpha + \beta = \dfrac{ – 4}{k}α+β=

   k

   −4

   Product of zeroes

   \pmb{ \alpha \beta = \dfrac{c}{a}}

   αβ=

   a

   c

   αβ=

   a

   c

   \sf\alpha \beta = \dfrac{4}{k}αβ=

   k

   4

   Apply Identity

   { \bigg( \alpha + \beta \bigg)}^{2} = { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2}(α+β)

   2

   =α

   2

   +2αβ+β

   2

   \sf { \bigg(\dfrac{ – 4}{k} } \bigg)^{2} = 24 + 2 \dfrac{4}{k}(

   k

   −4

   )

   2

   =24+2

   k

   4

   \sf\dfrac{16}{ {k}^{2} } = 24 + 8k

   k

   2

   16

   =24+8k

   \sf \: 16 = 24 {k}^{2} + 8k16=24k

   2

   +8k

   \sf \: \dfrac{16}{8} = 24k {}^{2} + k

   8

   16

   =24k

   2

   +k

   \sf \: 2 = 24 {k}^{2} + k2=24k

   2

   +k

   \sf \: 3 {k}^{2} + k – 2 = 03k

   2

   +k−2=0

   \sf \: 3 {k}^{2} + (3k – 2k) – 2 = 03k

   2

   +(3k−2k)−2=0

   \sf \: (3k – 2) \: or \: (k + 1)(3k−2)or(k+1)

   k = 2/3

   Or

   k = -1

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