Question

Which of the following has no real roots ?

x² + 5x + 6 = 0

x² + 3x + 5 = 0

x² - 3x - 5 = 0

x² - 5x - 6 = 0

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Answer 1

☯ GIVEN :

• 1) $\sf{x ^2 + 5x + 6 = 0}$

• 2)$\sf{x ^2 + 3x + 5= 0}$

• 3) $\sf{x ^2 - 3x - 5 = 0}$

• 4) $\sf{x ^2 - 5x - 6 = 0}$

☯TO FIND :

• Which of the following has no real roots?

☯ FORMULA REQUIRED :

• $\bigstar{\underline{ \boxed{ \red{ \bf{D = b^2 - 4ac }}}}}\bigstar$

➲SOLUTION :

✞ In First Question :

• a = 1 , b = 5 and c = 6.

Substituting the values :

$: \leadsto {\tt{D = b^2 - 4ac }}$

$: \leadsto{ \tt{D = (5)^2 - 4 \times1 \times 6}}$

$: \leadsto {\tt{D = 25 - 24}}$

$:\leadsto {\huge{\boxed{ \pink{\mathfrak{D = 1}}}}}$

$\huge {\therefore}$ Here, D is greater than 0 so roots are real and unequal.

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✞ In Second Question :

• a = 1 , b = 3 and c = 5.

Substituting the values :

$: \leadsto {\tt{D = b^2 - 4ac}}$

$: \leadsto {\tt{D = (3)^2 - 4 \times 1 \times 5}}$

$: \leadsto {\tt{D = 9 - 20}}$

$: \leadsto {\huge{\boxed{ \pink{\mathfrak{D = - 11}}}}}$

$\huge {\therefore}$ Here, D is smaller than 0 so roots are unreal and imaginary.

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✞ In Third Question :

• a = 1 , b = -3 and c = - 5.

Substituting the values :

$: \leadsto {\tt{D = b^2 - 4ac }}$

$: \leadsto {\tt{D = (-3)^2 - 4 \times 1 \times (-5)}}$

$: \leadsto {\tt{D = 9 + 20}}$

$:\leadsto{\huge{\boxed{\pink{\mathfrak{D = 29}}}}}$

$\huge {\therefore}$ Here, D is greater than 0 so roots are real and unequal.

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✞ In Fourth Question :

• a = 1 , b = -5 and c = - 6.

Substituting the values :

$: \leadsto {\tt{D = b^2 - 4ac }}$

$: \leadsto{\tt{D = (-5)^2 - 4 \times 1 \times (-6)}}$

$: \leadsto{\tt{D = 25 + 24}}$

$:\leadsto{\huge{\boxed{ \pink{\mathfrak{D = 49}}}}}$

$\huge {\therefore}$ Here, D is greater than 0 so roots are real and unequal.

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