Question

discuss the nature of the roots of the equation x^2+7x+12 =0 ​

Answer 1

Question :-

$\red{\sf{discus \: the \: nature \: of \: roots \: of \: the \:}} \\ \green{ \sf{ equation \: {x}^{2} + 7x + 12 = 0}}$

Answer :-

→Roots are real

→ Roots are -4 and -3 .

Formula used :-

To check about nature of roots of any quadratic equation we have to find discriminate of that equation.

$\sf{ \red{discriminate} \: = \pink{ {b}^{2} - 4ac}}$

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Explanation :-

Given equation is ,

$\rightarrow \sf{{x}^{2} + 7x + 12 = 0} \: \: \: \: \: ......(1)$

We know that standard equation ,

$\rightarrow \sf{ a{x}^{2} + bx + c = 0} \: \: \: \: ........(2)$

Comparing eq.(1) and (2) ,

We get,

$\implies \sf{ a \: = 1 \:, \: b \: = 7, \: \: c \: = 12}$

Now using the given formula,

$\implies \: \sf{{b}^{2} - 4ac} \\ \\ \implies \: {(7)}^{2} - 4 \times 1 \times 12 \\ \\ \implies \: 49 - 48 \\ \\ \sf{ discriminate \: = 1}$

Discriminate (d) > 0 ,

Therefore the roots of this equation are real .

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To finding the roots of this equation ,

Solve by factorisation method ,

$\implies \sf{ {x}^{2} + 4x + 3x + 12 = 0} \\ \\ \sf{\implies \: x(x + 4) + 3(x + 4) = 0} \\ \\ \implies \: \sf{(x + 4)( x + 3) = 0} \\ \\ case \: (1) \\ \\ \rightarrow \sf{ x + 4 = 0 }\\ \\ \rightarrow \boxed{\sf{ x \: = - 4}} \\ \\ case \: (2) \\ \\ \rightarrow \sf{ x + 3 = 0} \\ \\ \rightarrow \boxed{\sf{ x \: = - 3}}$

Roots of this equation are -4 and -3.

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