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Question: Solve the equation \({x^2} + 6x - 10 = 0\)...

Solve the equation x2+6x−10=0{x^2} + 6x - 10 = 0

Explanation

Solution

In the given question we are asked to solve the quadratic equation, that is we need to find the roots of the given quadratic equation. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} . Let ax2+bx+c=0a{x^2} + bx + c = 0 be a quadratic equation, by taking the coefficient of x2{x^2} as aa , and coefficient of xx as b and constant as cc .

Complete answer:
It is given that x2+6x−10=0{x^2} + 6x - 10 = 0 we aim to solve this equation that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} where aa - coefficient of the term x2{x^2}, bb - coefficient of the term xx, and cc - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation x2+6x−10=0{x^2} + 6x - 10 = 0, we have a=1a = 1, b=6b = 6, and c=−10c = - 10.
On substituting these terms in the formula, we get
x=−(6)±(6)2−4(1)(−10)2(1)x = \dfrac{{ - \left( 6 \right) \pm \sqrt {{{\left( 6 \right)}^2} - 4\left( 1 \right)\left( { - 10} \right)} }}{{2\left( 1 \right)}}
On simplifying this we get
x=−6±36+402x = \dfrac{{ - 6 \pm \sqrt {36 + 40} }}{2}
On further simplification we get
x=−6±762x = \dfrac{{ - 6 \pm \sqrt {76} }}{2} and thus we get x=−6±19×42⇒x=−6±2192x = \dfrac{{ - 6 \pm \sqrt {19 \times 4} }}{2} \Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt {19} }}{2}
x=−6±2192x = \dfrac{{ - 6 \pm 2\sqrt {19} }}{2}
⇒x=−6+2192\Rightarrow x = \dfrac{{ - 6 + 2\sqrt {19} }}{2} and x=−6−2192x = \dfrac{{ - 6 - 2\sqrt {19} }}{2}
On solving the above, we get
x=−3+19x = - 3 + \sqrt {19} and x=−3−19x = - 3 - \sqrt {19}
Thus, we got the roots of the given quadratic equation that is x=−3+19x = - 3 + \sqrt {19} and x=−3−19x = - 3 - \sqrt {19}

Note:
The possible values of the unknown variable in the equation is known as the roots of the equation.
The number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation.
If a=0a = 0 is it will form linear equation. Hence it is not possible.