# If x=16 is one of the root of the quadratic equation x^{2}-2\left(7x+16\right)=0,find the other...

## Question:

If {eq}x = 16 {/eq} is one of the root of the quadratic equation {eq}x^{2} - 2\left(7x + 16\right) = 0, {/eq} find the other root.

## Quadratic Equation and its Roots:

{eq}\\ {/eq}

Any quadratic equation has two roots or solutions, they may be real or complex in nature depending on the value of coefficients. Let us assume that we have two roots {eq}(\alpha) {/eq} and {eq}(\beta) {/eq} of the standard quadratic equation. Then there are two properties related to the sum and product of roots which are quite useful when we know anyone of the root and we have to determine the other one directly without using the standard formula.

\begin{align} &ax^{2} + bx + c = 0 \\[0.2cm] &x = \alpha, \; \beta \\[0.2cm] &\alpha + \beta = - \biggr( \dfrac {b}{a} \biggr) \\[0.2cm] &\alpha \times \beta = \biggr( \dfrac {c}{a} \biggr) \end{align}

{eq}\\ {/eq}

The described quadratic equation is given below:

\begin{align} x^{2} - 2 (7x + 16) &= 0 \\[0.2cm] x^{2} - 14x - 32 &= 0 &\left(\text{ Equation 1 } \right) \end{align}

We all know the standard form of a quadratic equation and its sum of roots property that is already mentioned in the context section:

\begin{align} ax^{2} + bx + c &= 0 &\left(\text{ Equation 2 } \right) \end{align}

Now, compare the coefficients of equations {eq}1{/eq} and {eq}2{/eq}:

$$a = 1, \; b - 14, \; c = -32$$

Let us assume that we have two roots of the given equation {eq}(\alpha \; \text {and} \; \beta) {/eq}:

\begin{align} \alpha + \beta &= - \biggr( \dfrac {b}{a} \biggr) \\[0.2cm] \alpha + 16 &= - \biggr( \dfrac {-14}{1} \biggr) \\[0.2cm] \alpha + 16 &= 14 \\[0.2cm] \alpha &=\boxed {-2 } \end{align}

Hence, the other root of the equation is {eq}x = -2 {/eq}.