Find the zeros of the given quadratic polynomial p(x) =x^2-2x-8​

2 thoughts on “Find the zeros of the given quadratic polynomial p(x) =x^2-2x-8​”

1. Required Answer:-

   Given:

   • p(x) = x² – 2x – 8

   To Find:

   • The zeros of the given equation.

   Solution:

   Given,

   → p(x) = x² – 2x – 8

   → x² – 2x – 8 = 0

   → x² – (4 – 2)x – 8 = 0

   → x² – 4x + 2x – 8 = 0

   → x(x – 4) + 2(x – 4) = 0

   → (x + 2)(x – 4) = 0

   By Zero-Product rule,

   → (x + 2) = 0 or (x – 4) = 0

   → x = -2, 4

   ★ So, the zeros of the given equation are -2 and 4.

   Answer:

   • x = -2, 4.

   •••♪

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2. Required Answer :

   • The value of x = –2 or 4.

   Concept :

   ☆ The given quadratic equation can be solved by two ways :-

   • By Prime factorisation – We’ll make the common factors of the middle term and then take a number as common to find the value of x. This method is known as splitting the middle term.
   • By quadratic formula – In this formula, we’ll find out the discriminant of the equation using formula b²–4ac, and then put it in the another formula to obtain the value of x. This method is known as quadratic formula.

   Step by step explanation :

   Given :

   • Quadratic equation = x²-2x-8

   To find :

   • Roots/zeroes of the quadratic equation.

   Solution :

   ⠀⠀⠀⠀⠀⠀⠀–––––––––––––First way–––––––––––

   ❥ By Prime factorisation,

   [tex]p(x) = {x}^{2} – 2x – 8[/tex]

   [tex]→[/tex] [tex]p(x) = {x}^{2} – 4x + 2x – 8[/tex]

   • Taking x and (+2) as common.

   [tex]→[/tex] [tex]p(x) = x(x – 4) + 2(x – 4)[/tex]

   [tex]→[/tex] [tex](x – 4)(x + 2)[/tex]

   • Now, for finding zeroes, put p(x) = 0

   ∴ [tex]p(x) = 0[/tex]

   [tex]→[/tex] [tex](x – 4)(x + 2) = 0[/tex]

   [tex]→[/tex] [tex](x – 4) = 0 \: or \: (x + 2) = 0[/tex]

   [tex]→[/tex] [tex]x = 4 \: or \: x = – 2[/tex] ✔️

   ⠀⠀ ⠀⠀—————–Second way————–

   ❥By quadratic formula,

   Given, [tex]p(x) = {x}^{2} – 2x – 8[/tex]

   • On comparing p(x) with ax²+bx+c, we get

   1. a = 1
   2. b = -2
   3. c = -8

   • Finding Discriminant using the formula, b²-4ac.

   [tex] {b}^{2} – 4ac[/tex]

   [tex]→[/tex] [tex]( – 2)^{2} – 4( 1)( – 8)[/tex]

   [tex]→[/tex] [tex]4 + 32[/tex]

   [tex]→[/tex] [tex]36[/tex] ——(i)

   • Now, using the quadratic formula including the value of discriminant for finding the roots.

   [tex] \frac{ – b± \sqrt{b^{2} – 4ac } }{2a} [/tex]

   [tex]→[/tex] [tex] \frac{ +2 ± \sqrt{36 } }{2} [/tex] [ from (i) ]

   [tex]→[/tex] [tex] \frac{2 – 6}{2} \: or \: \frac{2 + 6}{2} [/tex]

   [tex]→[/tex] [tex] \frac{ – 4}{2} \: or \: \frac{8}{2} [/tex]

   ∴[tex]∴ \: x = – 2 \: or \: 4[/tex] ✔️

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