Question

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

Answer 1

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,

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

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

$\tt: \implies x = 2.$

Answer 2

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;

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

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