Math, asked by rajaninsan352pc3ytw, 1 year ago

find the zeroes of the following quadratic equation

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Answers

Answered by shashankavsthi
1
I m using mid term splitting method.

 {x}^{2}  - 2x - 8 = 0 \\  {x}^{2}  - 4x + 2x - 8 = 0 \\ x(x - 4) + 2(x - 4) = 0 \\ (x + 2)(x - 4) = 0 \\ so \: thezeroes \: of \: equations \: are \\  \\ x =  - 2 \: and \: x = 4
hope it will help u
Answered by Anonymous
3

Question :

  • find the zeroes of the quadratic equation x² - 2x - 8

Given :

 \tt \implies  {x}^{2}  - 2x - 8

To Find :

  • Zeroes of the quadratic equation

Solution :

Quadratic formula

 \large \boxed{ \tt x =  \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}  } \\  \\ \tt \implies  {x}^{2}  - 2x - 8 \\ \\  \tt here \\   \\ \tt \boxed{ \tt a = 1} \:  \:  \: \boxed{ \tt b =  - 2 } \:  \:  \:  \boxed{ \tt c =  - 8} \\  \\  \tt Put \: values \: in \: the \: formula \\  \\ \tt x =  \frac{ - ( - 2) \pm \sqrt{ { (- 2)}^{2} - 4  \times 1 \times ( - 8) } }{2 \times 1} \\  \\  \tt x =   \frac{2 \pm \sqrt{4  +  32} }{2}  \\  \\  \tt x =  \frac{2 \pm  \sqrt{36} }{2}  \\  \\  \tt x =  \frac{2 \pm 6}{2}  \\  \\   \tt \bullet \: x =  \frac{2 +  6}{2}  \\  \\ \large\boxed{ \tt x = 4} \\  \\  \tt \bullet \: x = \frac{2 - 6}{2}  \\  \\\large \boxed{ \tt x =  - 2}

 \large \underline{\tt Zeroes \: of \: equation \: are \: 1 \: and \:  - 2}

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