solve the quadratic equation x^2-3√5x+10=0 AND -x^2+7x-10=0​

Question

solve the quadratic equation x^2-3√5x+10=0 AND -x^2+7x-10=0​

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Cora 6 days 2021-10-11T08:13:03+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-11T08:14:05+00:00

    SOLUTION :

    \bigstar Firstly, we know that formula of the quadratic equation :

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

    We have;

    • x² – 3√5x + 10 = 0
    • -x² + 7x – 10 = 0

    Now,

    Given equation will compared with ax² + bx + c = 0

    • a = 1
    • b = –3√5
    • c = 10

    Substitute the given values,we get;

    \longrightarrow\sf{x=\dfrac{-(-3\sqrt{5} )\pm\sqrt{(-3\sqrt{5})^{2} -4\times 1\times 10 } }{2\times 1} }\\\\\\\longrightarrow\sf{x=\dfrac{3\sqrt{5} \pm\sqrt{9\times 5-40} }{2} }\\\\\\\longrightarrow\sf{x=\dfrac{3\sqrt{5} \pm\sqrt{45-40} }{2} }\\\\\\\longrightarrow\sf{x=\dfrac{3\sqrt{5}\pm\sqrt{5}  }{2} }\\\\\\\longrightarrow\sf{x=\dfrac{3\sqrt{5} +\sqrt{5} }{2} \:\:\:x=\dfrac{3\sqrt{5} -\sqrt{5} }{2} }\\\\\\\longrightarrow\sf{x=\dfrac{\sqrt{5} (3+1)}{2} \:\:Or\:\:x=\dfrac{\sqrt{5}(3-1) }{2} }\\\\\\

    \longrightarrow\sf{x=\dfrac{\sqrt{5} \times \cancel{4}}{\cancel{2}} \:\:\:Or\:\:\:x=\dfrac{\sqrt{5} \times\cancel{ 2}}{\cancel{2}} }\\\\\\\longrightarrow\bf{x=2\sqrt{5} \:\:Or\:\:x=\sqrt{5} }

    Again,

    • a = –1
    • b = 7
    • c = –10

    Substitute the given values,we get;

    \longrightarrow\sf{x=\dfrac{-7\pm\sqrt{(7)^{2} -4\times (-1)\times (-10)} }{2\times (-1)} }\\\\\\\longrightarrow\sf{x=\dfrac{-7\pm\sqrt{49-4\times 10} }{-2} }\\\\\\\longrightarrow\sf{x=\dfrac{-7\pm\sqrt{49-40} }{-2} }\\\\\\\longrightarrow\sf{x=\dfrac{-7\pm\sqrt{9} }{-2} }\\\\\\\longrightarrow\sf{x=\dfrac{-7\pm3}{-2} }\\\\\\\longrightarrow\sf{x=\dfrac{-7+3}{-2} \:\:or \:\:x=\dfrac{-7-3}{-2} }\\\\\\

    \longrightarrow\sf{x=\cancel{\dfrac{-4}{-2}} \:\:\:Or\:\:\:x=\cancel{\dfrac{-10}{-2} }}\\\\\\\longrightarrow\bf{x=2\:\:\:Or\:\:\:x=5}

    0
    2021-10-11T08:14:23+00:00

    SolutioN :

    We have, Quadratic Equation.

     \tt \dagger \:  \:  \:  \:  \:  {x}^{2}  - 3 \sqrt{5} x + 10 = 0.

     \tt \dagger \:  \:  \:  \:  \:  -  {x}^{2}  + 7  x  -  10 = 0.

    Let’s try with Quadratic Formula.

     \dagger \:  \:  \:  \:  \:  \boxed{ \tt{x =  \dfrac{ - b \pm  \sqrt{ {b}^{2}  - 4ac} }{2a}} }

         \tt: \implies x =  \dfrac{ 3 \sqrt{5}  \pm  \sqrt{ {(3 \sqrt{5}) }^{2}  - 4 \times 1  \times 10} }{2}

         \tt: \implies x =  \dfrac{ 3 \sqrt{5}  \pm  \sqrt{ {45}- 4 0} }{2}

         \tt: \implies x =  \dfrac{ 3 \sqrt{5}  \pm  \sqrt{ {5}} }{2}

         \tt: \implies x =  \dfrac{  \sqrt{5}  (3\pm{ 1} )}{2}

    Either,

         \tt: \implies x =  \dfrac{  \sqrt{5}  (3 + { 1} )}{2}

         \tt: \implies x =  \dfrac{ 4 \sqrt{5}  }{2}

         \tt: \implies x =  2\sqrt{5}.

    Or,

         \tt: \implies x =  \dfrac{  \sqrt{5}  (3 - { 1} )}{2}

         \tt: \implies x =  \sqrt{5} .

    ( ii )

    • a = 1.
    • b = – 7.
    • c = 10.

         \tt: \implies x =  \dfrac{ 7  \pm  \sqrt{ {( - 7) }^{2}  - 4 \times 1  \times 10} }{2}

         \tt: \implies x =  \dfrac{ 7  \pm  \sqrt{ {49}  - 40} }{2}

         \tt: \implies x =  \dfrac{ 7  \pm  \sqrt{ 9} }{2}

         \tt: \implies x =  \dfrac{ 7  \pm3}{2}

    Either,

         \tt: \implies x =  \dfrac{ 7   + 3}{2}

         \tt: \implies x =  \dfrac{ 10}{2}

         \tt: \implies x =  5.

    Or,

    \tt: \implies x =  \dfrac{ 7  - 3}{2}

         \tt: \implies x =  \dfrac{4}{2}

         \tt: \implies x = 2.

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