Question

solve the following quadratic equation by using formula method . 5x²-13+ 8 =0​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:x\:=\:\dfrac{8}{5}}}\sf\:\quad\:OR\:\quad\:\boxed{\red{\sf\:x\:=\:1\:}}}$

Step-by-step-explanation:

The given quadratic equation is 5x² - 13x + 8 = 0.

We have to find the roots of the quadratic equation by using formula method.

Now, comparing the given equation with ax² + bx + c = 0, we get,

5x² - 13x + 8 = 0

• a = 5
• b = - 13
• c = 8

Now, we know that,

$\displaystyle{\pink{\sf\:x\:=\:\dfrac{-\:b\:\pm\:\sqrt{b^2\:-\:4ac}}{2a}}\sf\:\qquad\cdots[\:Quadratic\:formula\:]}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{-\:(\:-\:13\:)\:\pm\:\sqrt{(\:-\:13\:)^2\:-\:4\:\times\:5\:\times\:8}}{2\:\times\:5}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{13\:\pm\:\sqrt{169\:-\:4\:\times\:40}}{10}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{13\:\pm\:\sqrt{169\:-\:160}}{10}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{13\:\pm\:\sqrt{9}}{10}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{13\:\pm\:3}{10}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{13\:+\:3}{10}\:\quad\:OR\:\quad\:x\:=\:\dfrac{13\:-\:3}{10}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\cancel{16}}{\cancel{10}}\:\quad\:OR\:\quad\:x\:=\:\cancel{\dfrac{10}{10}}}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:x\:=\:\dfrac{8}{5}}}}\sf\:\quad\:OR\:\quad\:\underline{\boxed{\red{\sf\:x\:=\:1\:}}}}$

Answer 2

Question :-

solve the following quadratic equation by using formula method . 5x²-13x + 8 =0

Solution :-

Given :

• 5x² - 13x + 8 = 0

Need to find :

• value of x

Using formula :

$\rm:\longmapsto{x = \frac{ - b \pm \sqrt{b {}^{2} - 4ac} }{2a} }$

Comparing the given equation with ax² + bx + c = 0.

• a = 5
• b = -13
• c = 8

On using formula,

$\rm:\longmapsto{ x = \frac{ - ( - 13) \pm \sqrt{( - 13) {}^{2} - 4(5)(8)} }{2(5)} } \\ \\ \rm:\longmapsto{x = \frac{13 \pm \sqrt{169 - 160} }{10} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now,

$\small\rm:\longmapsto{x = \frac{13 + \sqrt{169 - 160} }{10} \: \: \: or \: \: \: x = \frac{13 - \sqrt{169 - 160} }{10} }$

Then,

$\rm:\longmapsto{x = \frac{13 + \sqrt{9} }{10} \: \: \: or \: \: \: x = \frac{13 - \sqrt{9} }{10} } \\ \\\rm:\longmapsto{x = \frac{13 + 3}{10} \: \: \: or \: \: \: x = \frac{13 - 3}{10} } \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{x = \frac{16}{10} \: \: \: or \: \: \: x = \frac{10}{10} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto \boxed {\rm{x = \frac{8}{5} \: \: \: or \: \: \: x = 1 }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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