Question

$\tt\huge\underline{Question}:-$


$\sf \: Find \: the \: roots \: of \: the\: following \\\sf \: quadratic \: equations \: by \: using \\ \sf\underline{QUADRATIC \: FORMULA \: }method.$


a) 7x² + 6x + 1 = 0

b) 3x² - 10x + 8 = 0​

Answer 1

Answer:

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Answer 2

Solution :

a) 7x² + 6x + 1 = 0

The given equation is

$\sf{} \implies \: 7x^{2}+ 6x + 1 = 0$

Comparing it with ax² + bx + c = 0, we get,

a = 7 , b = 6 , c = 1

By finding Discriminant;

$\: \: \: \: \: \: \: \: \: \: \star \boxed{\sf{ \bold{D = {b}^{2} - 4ac}}} \star$

By substituting values;

$\sf{} \rightarrow \: D =(6 {)}^{2} - 4(7)(1) \\ \\ \sf{} \rightarrow \: D =36 - 28(1) \\ \\ \sf{} \rightarrow \: D =36 - 28 \\ \\ \sf{} \rightarrow \: D =8 \\ \\ \sf{} \rightarrow \: \sqrt{D} = \sqrt{8} \\ \\ \sf{} \rightarrow \: \sqrt{D} = 2\sqrt{2}$

Therfore, since the D = 8 which implies that,

$\implies \: \sf{} D \: > 0$

Hence, the roots are real and different.

Now let us find the roots using Quadratic Formula ;

$\: \: \: \: \: \: \boxed{\sf{} { \bold{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

or

$\: \: \: \: \: \: \boxed{\sf{} { \bold{x = \frac{ - b \pm \sqrt{D} }{2a} }}}$

By putting values

$\sf{} \implies \: x = \dfrac{ - 6 \pm \: 2\sqrt{2} }{2(7)} \\ \\ \sf{} \implies \: x = \dfrac{ - 6 \pm 2\sqrt{2}}{14} \\ \\ \sf{} \implies \: x = \dfrac{ - 6 + 2\sqrt{2}}{14} \: \: \: \: \dfrac{ - 6 - 2\sqrt{2}}{14} \\ \\ \sf{} \implies \: x = \dfrac{ - 3+2\sqrt{2}}{7} \: \: \: \dfrac{ 3-2\sqrt{2}}{7} \\ \\ \sf{} \implies \: x = \dfrac{3-2\sqrt{2}}{ 7} \: \: \: \: \dfrac{ 3-2\sqrt{2}}{7}$

Therefore, roots are 3-2√2/7 and 3-2√2/7

b) 3x² - 10x + 8 = 0

The given equation is

$\sf{} \implies \: 3x^{2} - 10x + 8 = 0$

Comparing it with ax² + bx + c = 0, we get,

a = 3, b = -10 , c = 8

By finding Discriminant;

$\: \: \: \: \: \: \: \: \: \: \star \boxed{\sf{ \bold{D = {b}^{2} - 4ac}}} \star$

By substituting values;

$\rightarrow \sf{} \: D =( - 10 {)}^{2} - 4(3)(8) \\ \\ \rightarrow \sf{} \: D =100 - 12(8) \\ \\ \rightarrow \sf{} \: D =100 - 96 \\ \\ \rightarrow \sf{} \: D =4 \\ \\ \rightarrow \sf{} \: \sqrt{} D = \sqrt{4} \\ \\ \rightarrow \sf{} \: \sqrt{} D =2$

Therfore, since the D = 2 which implies that,

$\implies \: \sf{} D \: > 0$

Hence, the roots are real and different.

Now let us find the roots using Quadratic Formula ;

$\: \: \: \: \: \: \boxed{\sf{} { \bold{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

or

$\: \: \: \: \: \: \boxed{\sf{} { \bold{x = \frac{ - b \pm \sqrt{D} }{2a} }}}$

By putting values ;

$\sf{} \implies \: x = \dfrac{ - 10 \pm \: 2}{2(3)} \\ \\ \sf{} \implies \: x = \dfrac{ - 10 \pm \: 2}{6} \\ \\ \sf{} \implies \: x = \dfrac{ - 10 + 2}{6} \: \: \: \dfrac{ - 10 - 2}{6 } \\ \\ \sf{} \implies \: x = \dfrac{ - 8}{2} \: \: \: \dfrac{ - 12}{6} \\ \\ \sf{} \implies \: x = \dfrac{ - 4}{1} \: \: \: \dfrac{ - 2}{1}$

Therefore, roots are -4 and -2.