Mathematics, 08.01.2021 16:10 yiyirojastorres

Solve 2x^2 + x - 15 = 0

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Answer from: logan541972

Step-by-step explanation:

$\tt{2 {x}^{2} + x - 15 = 0}$

USING THE QUADRATIC FORMULA :

Here ,

a = 2 , b = 1 and c = - 15

Substitute the values into the quadratic equation :

⇾ $\tt{x \: = \: \frac{ - b \: ± \: \sqrt{ {b}^{2} - 4ac} \: }{2a} }$

⇾ $\tt{x = \frac{ - 1\: ± \: \sqrt{ {1}^{2} - 4 \times 2 \times ( - 15) } }{2 \times 2}}$

Simplify

⇾ $\tt{x = \frac{ -1 \:± \sqrt{1 - 8 \times ( - 15)} }{4}}$

⇾ $\tt{x = \frac{ - 1\:± \: \sqrt{1 + 120} }{4}}$

⇾ $\tt{x = \frac{ - 1 \:± \: \sqrt{121} }{4}}$

⇾ $\tt{x = \frac{ - 1\: ± \: \sqrt{ {11}^{2} } }{4}}$

⇾ $\tt{x = \frac{ - 1 \:± \: 11 }{4}}$

Now , we can split this into two answers because of the plus minus ( ± ) symbol.

Taking positive ( + ) sign :

⇾ $\tt{x = \frac{ - 1+ 11}{4}}$

⇾ $\tt{x = \frac{10}{4}}$

⇾ $\boxed{ \tt{x = \frac{5}{2} }}$

Again , taking negative ( - ) sign :

⇾ $\tt{x = \frac{ - 1 - 11}{4}}$

⇾ $\tt{x = \frac{ - 12}{4}}$

⇾ $\boxed{ \tt{x = - 3}}$

☥ $\red{ \bold{ \boxed{ \boxed{ \tt{Our \: final \: answer : x = \frac{5}{2} \: or \: - 3}}}}}$

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$\underline{ \underline{ \sf{Explore \: more !!}}} :$

There are various methods of solving quadratic equations. They are as follows :

Solving a quadratic equation by factorisation method : In this method , the second order of polynomial ax² + bx + c is factorised and expressed as the product of two linear factors. Then each linear factors is separately solved to get the required solutions of the equation by applying zero factor property. In zero factor property , if p • q = 0 then either p = 0 or q = 0 .In other words , if the product of two numbers is 0 , then one or both of the numbers must be zero.Solving a quadratic equation by completing the square : In this method , we transpose the constant term ( c ) to R.H.S , then L.H.S to expressed as perfect square expression.Solving a quadratic equation by using formula : In solving a quadratic equation of the form ax² + bx + c = 0 by completing the square , we obtain two roots of x , which are : $\sf{ \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a}}$ and $\sf{ \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}}$ . These roots can be written as $\sf{x = \frac{ - b \: ± \: \sqrt{ {b}^{2} - 4ac} }{2a}}$ where a is the coefficient of x² , b is the coefficient of x and c is the constant term. We use this formula to find the required solutions of the given quadratic equation.

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Answer from: Quest

Where's the rest of the

Answer from: Quest

step-by-step explanation:

only stephen hawking would know.

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