
How do you solve $ {x^2} + 5x - 2 = 0 $ ?
Answer
546.3k+ views
Hint: When the degree of a polynomial equation is two, the polynomial equation is called a quadratic equation. The values of the unknown quantity at which the value of the whole quadratic equation comes out to be zero are called the zeros/factors/solutions/roots of that given equation. Thus by solving the equation, we mean to find the roots of the given equation.
Complete step-by-step answer:
We have to solve $ {x^2} + 5x - 2 = 0 $
On comparing this equation with the standard quadratic equation
$ a{x^2} + bx + c = 0 $ , we get – $ a = 1,\,b = 5\,,c = - 2 $ .
We cannot solve this equation by factorization, so we use the quadratic formula –
$
x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {{{(5)}^2} - 4 \times 1 \times ( - 2)} }}{{2(1)}} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 + 8} }}{2} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {33} }}{2} \;
$
Hence when $ {x^2} + 5x - 2 = 0 $ , we get $ x = \dfrac{{ - 5 + \sqrt {33} }}{2}\,and\,x = \dfrac{{ - 5 - \sqrt {33} }}{2} $ .
So, the correct answer is “ $ x = \dfrac{{ - 5 + \sqrt {33} }}{2}\,and\,x = \dfrac{{ - 5 - \sqrt {33} }}{2} $ ”.
Note: For solving a question by factorization, we have to express the coefficient of x as the sum of two numbers such that the product of these two numbers must be equal to the product of the coefficient of $ {x^2} $ and the constant. Let $ {b_1} + {b_2} = b $ , so the condition for factorization is that $ a \times c = {b_1} \times {b_2} $ . When an equation cannot be factorized, we solve it by another method called the quadratic formula as in this question.
Complete step-by-step answer:
We have to solve $ {x^2} + 5x - 2 = 0 $
On comparing this equation with the standard quadratic equation
$ a{x^2} + bx + c = 0 $ , we get – $ a = 1,\,b = 5\,,c = - 2 $ .
We cannot solve this equation by factorization, so we use the quadratic formula –
$
x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {{{(5)}^2} - 4 \times 1 \times ( - 2)} }}{{2(1)}} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 + 8} }}{2} \\
\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {33} }}{2} \;
$
Hence when $ {x^2} + 5x - 2 = 0 $ , we get $ x = \dfrac{{ - 5 + \sqrt {33} }}{2}\,and\,x = \dfrac{{ - 5 - \sqrt {33} }}{2} $ .
So, the correct answer is “ $ x = \dfrac{{ - 5 + \sqrt {33} }}{2}\,and\,x = \dfrac{{ - 5 - \sqrt {33} }}{2} $ ”.
Note: For solving a question by factorization, we have to express the coefficient of x as the sum of two numbers such that the product of these two numbers must be equal to the product of the coefficient of $ {x^2} $ and the constant. Let $ {b_1} + {b_2} = b $ , so the condition for factorization is that $ a \times c = {b_1} \times {b_2} $ . When an equation cannot be factorized, we solve it by another method called the quadratic formula as in this question.
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