Question

Solve
2x + 5 = x²

Thanks in advance !

Answer 1

Answer:

= 1 ± /6

try it in steps to get full

Answer 2

$\implies\sf2x + 5 = {x}^{2} \\ \\ \implies\sf {x}^{2} - 2x - 5$

• we have to find the Discriminant of this equation by using the formula:-

$\longrightarrow\underline\green{\boxed{\bf D = b²-4ac}}$

$\implies\sf D = ( - 2) ^{2} - 4(1)( - 5) \\ \\ \implies\sf D = 4 + 20 = 24$

• since, D>0 hence, this equation will have distinct and real roots.

• Formula to be applied now is as follows:-

$\longrightarrow\underline\pink{\boxed{\bf x = \dfrac{-b±√D}{2a}}}$

$\implies\sf x = \dfrac{ - ( - 2)± \sqrt{24} }{2(1 )}\\ \\ \implies\sf x = \frac{4±2 \sqrt{6} }{2}$

• Now, when X = $\sf\dfrac{4+2√6}{2 }$

$\implies \sf x = \dfrac{4 + 2 \sqrt{6} }{2} \\ \\ \implies \sf \: x = \frac{ \cancel2(2 + \sqrt{6} )}{ \cancel2} \\ \\ \implies\boxed {\purple{\sf x = 2 + \sqrt{6}}}$

• when X = $\sf\dfrac{4-2√6}{2 }$

$\implies \sf x = \dfrac{4 - 2 \sqrt{6} }{2} \\ \\ \implies\sf \: x = \frac{ \cancel2(2 - \sqrt{6} )}{ \cancel2} \\ \\ \implies\boxed {\purple{\sf x = 2 - \sqrt{6}}}$

hence, the required zeros are:-

$\underline\orange{\boxed{\bf x = (2+√6) \:and \:(2-√6)}}$