Question

if one root of the equation x2+5x-14=0 is 2 ,then find the other root ​

Answer 1

Given-

$\rm {\bf{{x}^{2} + 5x - 14 = 0}} \\ \bf \: is \: a \: quadratic \: equation \: \\ \rm one \: root = 2 \: or \: \alpha = 2$

To Find-

$\rm \: second \: root \: or \: \beta$

Concept used-

$\bf \rightarrow\alpha + \beta = \dfrac{ - b}{a} \\ \\ \bf \rightarrow \alpha \beta = \dfrac{c}{a} \: \: \: \: \: \: \: \:$

Solution

$\rm {x}^{2} + 5x - 14 = 0$

Method-1st

Here,

• a=1

• b=5

• c=-14

We know,

$\rm\alpha + \beta = \dfrac{ - b}{a} \: \: \: \: \: \: \: \: \\ \\ \bf \: putting \: value \: \: \\ \\ \implies \rm2 + \beta = \frac{ - 5}{1} \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \beta = - 5 - 2 \: \: \: \: \: \: \: \: \: \: \: \\\ \\ \implies \underline{ \boxed{ \huge \bf \rm \beta = - 7}}$

Method- 2nd

We also know,

$\rm \: \alpha \beta = \dfrac{c}{a} \\ \\ \bf \: putting \: values \\ \\ \implies \rm \: 2 \beta = \frac{ - 14}{1} \\ \\ \rm \implies \: \beta = \frac{ - 14}{2} \\ \\ \rm \implies \underline{ \boxed{ \huge \rm \: \beta = - 7}}$

Hence,the second zero is-7

Method-3rd

$\rm \: {x}^{2} + 5x - 14 = 0 \: \: \: \: \: \\ \\ \bf \: by \: middle \:spliting \\ \\ \rm \implies \: {x}^{2} + 7x - 2x - 14 = 0 \: \: \\ \\ \implies \rm {x}(x + 7) - 2(x + 7) = 0 \\ \\ \implies \rm \: (x - 2)(x + 7) = 0 \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies \boxed{ \bf \: x = 2 \: or \: x = - 7 } \: \: \: \: \: \: \: \: \:$

Hence the first zero is 2 and second zero is-7

Note

If you're in class 9th or any other junior class then solve this question with method number third because both the above method were not in the syllabus of class 9th or 8th but if you're in class 10 it was compulsory to do with above 2methods