Question

solve the quadratic equation using formulae
$\sf\implies x^2 = 18x - 77$
​

Answer 1

✰ AnSwer:

• The value of x is 11 or 7.

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✒ GiVen :

• $\sf x^2 = 18x - 77$ is the given eqn.

✒ SoluTion :

Let us solve the eqn now,

$\sf\longmapsto \: \: x^2 = 18x - 77 \\ \\ \sf\longmapsto \: \: x^2 - 18x + 77 = 0$

In this above eqn we can observe the same form of eqn as,

☢⠀$\: \: {\boxed{\sf{ \: ax^2 + bx + c = 0 }}}$

Now, we know that, ⠀⠀⠀⠀

• a = 1
• b = -18
• c = 77

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Now, let us find the discriminant,

☢⠀$\: \: \: {\boxed{\purple{\large{\sf{x = \dfrac{ \: \: - b \pm \sqrt D }{2a \: \: }}}}}}$

☢⠀$\: \: \: {\boxed{\red{\large{\sf{ \: D = b^2 - 4ac \: }}}}}$

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By using the above formulas,

$\sf\longmapsto \: \: D = b^2 - 4ac$

$\sf\longmapsto \: \: D = (-18)^2 - 4 \times 1 \times 77$

$\sf\longmapsto \: \: D = 324 - 308$

$\sf\longmapsto \: \: { \underline{ \boxed{ \sf{ \blue{ \bold{ \: D = 16 \: }}}}}} \: \bigstar$

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Now, substitute the value of D in the above formula,

$\sf\longmapsto \: \: x = \dfrac{ -b \pm\sqrt D }{2a} \\ \\ \sf\longmapsto \: \: x = \dfrac{ -( -18) \pm\sqrt {16} }{2\times 1 } \\ \\ \sf\longmapsto \: \: x = \dfrac{ 18 \pm 4 }{2}$

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In this condition, there may be 2 values of x, that is,

$\sf \longmapsto \: \: x = \dfrac{ 18 + 4 }{ 2 } \\ \\ \sf\longmapsto x = \dfrac {22}{2} \\ \\\sf \longmapsto { \underline{ \boxed{ \pink{ \sf{ \bold{ \: x = 11 \: }}}}}} \: \bigstar$

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Similarly, 2nd value of x would be,

$\sf \longmapsto \: \: x = \dfrac{ 18 - 4 }{ 2 } \\ \\ \sf \longmapsto \: \: x = \dfrac {14}{2} \\ \\ \sf \longmapsto \: \: { \underline{ \boxed{ \red{ \sf{ \bold{ \: x = 7 \: }}}}}} \: \bigstar$

Therefore,

• x = 11 or x = 7

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Answer 2

Solution :

We have quadratic equation p(x) = x² = 18x - 77

Zero of the polynomial p(x) = 0

∴ x² - 18x + 77 = 0

As we know that given above equation compared with ax² + bx + c;

• a = 1
• b = -18
• c = 77

$\underline{\boldsymbol{Using\:by\:quadratic\:formula\::}}}$

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

$\longrightarrow\tt{x=\dfrac{-(-18)\pm\sqrt{(-18)^{2} -4\times 1\times 77} }{2\times 1} }\\\\\\\longrightarrow\tt{x=\dfrac{18\pm\sqrt{324-308} }{2} }\\\\\\\longrightarrow\tt{x=\dfrac{18\pm \sqrt{16} }{2}} \\\\\\\longrightarrow\tt{x=\dfrac{18\pm 4}{2} }\\\\\\\longrightarrow\tt{x=\dfrac{18 + 4}{2} \:\:\:Or\:\:\:x=\dfrac{18-4}{2} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{22}{2}} \:\:\:Or\:\:\:x=\cancel{\dfrac{14}{2}} }$

$\longrightarrow\bf{x=11\:\:\:Or\:\:\:x=7}$

Thus;

x = 11 or x = 7 are two roots of the polynomial .