Mathematics, 08.01.2021 16:10 yiyirojastorres

Solve 2x^2 + x - 15 = 0

Answers: 3

Show answers

Answers

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 :

⇾ $\tt{x = \frac{ - 1 - 11}{4}}$

⇾ $\tt{x = \frac{ - 12}{4}}$

⇾ $\boxed{ \tt{x = - 3}}$

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

⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆⋆

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

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.

Hope I helped! ツ

Have a wonderful day / night ! ♡

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

Answer from: Quest

50 % is the answer !

Answer from: Quest

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 18:30

The formula for the lateral area of a right cone is la = rs, where r is the radius of the base and s is the slant height of the cone.which are equivalent equations?

Answers: 3

Answer

Mathematics, 21.06.2019 18:30

Which representation has a constant of variation of -2.5

Answers: 1

Answer

Mathematics, 21.06.2019 19:30

What is the effect on the graph of the function f(x)=x when f(x) is replaced with -f(x)+4

Answers: 1

Answer

Mathematics, 21.06.2019 20:30

For f(x) = 3x+1 and g(x) = x2 - 6, find (g/f)(x)