How do you solve $4 + 7 {(3 x - 5)}^{2} = 1$ $4+7(3x-5)^2 =1$ ?

Question

1370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T15:03:53

$x_{1} = \frac{1}{21} (35 - i \sqrt{21})$ $x_1=1/21(35-isqrt(21))$
$x_{2} = \frac{1}{21} (35 + i \sqrt{21})$ $x_2=1/21(35+isqrt(21))$

Expanding after the frormula

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$
we get
$63 x^{2} - 210 x + 148 = 0$ $63x^2-210x+148=0$
dividing all by $63$ $63$

$x^{2} - \frac{210}{63} x + \frac{148}{63} = 0$ $x^2-210/63x+148/63=0$
solving this equation

$x_{1} = \frac{1}{21} (35 - i \sqrt{21})$ $x_1=1/21(35-isqrt(21))$

$x_{2} = \frac{1}{21} (35 + i \sqrt{21})$ $x_2=1/21(35+isqrt(21))$

where $i^{2} = - 1$ $i^2=-1$

How do you solve 4+7(3x−5)2=14+7(3x-5)^2 =1?