
How do you solve for $x$ in this equation ${x^2} + 38 = 3x - 12$ ?
Answer
537k+ views
Hint: In this question, $x$ is raised to the power 2 so we are given a quadratic equation in terms of x and we have to solve for the value of x. A quadratic equation is solved by using factorization or the quadratic formula or completing the square method. The given equation needs to be first converted to the standard form, that is, $a{x^2} + bx + c = 0$ and then we will compare the equation with the standard form and get the value of the coefficients a, b and c. Then we will put these values in the quadratic formula and solve it. As the degree of the equation is 2, so we will get 2 values of x.
Complete step-by-step solution:
We are given that ${x^2} + 38 = 3x - 12$
Rearranging the equation, we get –
$\Rightarrow {x^2} - 3x + 50 = 0$
On comparing it with $a{x^2} + bx + c = 0$ , we get –
$a = 1,\,b = - 3\,and\,c = 50$
The quadratic formula is given as –
$
\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4 \times 1 \times (50)} }}{{2(1)}} \\
\Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 200} }}{2} = \dfrac{{3 \pm \sqrt { - 191} }}{2} \\
\Rightarrow x = \dfrac{{3 \pm \sqrt {191} i}}{2}.....(i = \sqrt { - 1} ) \\
$
Hence when ${x^2} + 38 = 3x - 12$ , we get $x = \dfrac{{3 + \sqrt {191} i}}{2}$ and $x = \dfrac{{3 - \sqrt {191} i}}{2}$ .
Note: We get a polynomial equation when the unknown variable quantity in an algebraic expression is raised to some non-negative integer as the power, and the highest power of the unknown quantity in the equation is known as the degree of the polynomial equation. The given equation has a degree of 2 as the highest exponent of x in the given equation is 2, so the polynomial equation is called a quadratic equation. The given equation cannot be solved by factorization so we use the quadratic formula to find the values of x. We can verify that the answer obtained is correct by putting the obtained value of x in the given equation.
Complete step-by-step solution:
We are given that ${x^2} + 38 = 3x - 12$
Rearranging the equation, we get –
$\Rightarrow {x^2} - 3x + 50 = 0$
On comparing it with $a{x^2} + bx + c = 0$ , we get –
$a = 1,\,b = - 3\,and\,c = 50$
The quadratic formula is given as –
$
\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4 \times 1 \times (50)} }}{{2(1)}} \\
\Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 200} }}{2} = \dfrac{{3 \pm \sqrt { - 191} }}{2} \\
\Rightarrow x = \dfrac{{3 \pm \sqrt {191} i}}{2}.....(i = \sqrt { - 1} ) \\
$
Hence when ${x^2} + 38 = 3x - 12$ , we get $x = \dfrac{{3 + \sqrt {191} i}}{2}$ and $x = \dfrac{{3 - \sqrt {191} i}}{2}$ .
Note: We get a polynomial equation when the unknown variable quantity in an algebraic expression is raised to some non-negative integer as the power, and the highest power of the unknown quantity in the equation is known as the degree of the polynomial equation. The given equation has a degree of 2 as the highest exponent of x in the given equation is 2, so the polynomial equation is called a quadratic equation. The given equation cannot be solved by factorization so we use the quadratic formula to find the values of x. We can verify that the answer obtained is correct by putting the obtained value of x in the given equation.
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