How do you solve $4 x^{2} = 25$ $4x^2=25$ ?

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Answer 1 · 2016-04-17T04:55:09

$x = \pm \frac{5}{2}$ $x=+-5/2$

Use the $golden rule$ $color(gold)("golden rule"$ of Algebra

$What we do on one side,must be done on the other side also$ $"What we do on one side,must be done on the other side also"$

Our aim is to isolate $x$ $x$ to find its value

$4 x^{2} = 25$ $color(blue)(4x^2=25$

Divide both sides by $4$ $4$

$rarr(cancel4x^2)/cancel4=25/4$

$rarrx^2=25/4$

Take the square root of both sides

$rarrsqrt(x^2)=sqrt(25/4$

$rarrx=sqrt(25/4)$

Remember that,if we take the square root of a number,it can be positive or negative

So,

$rarrx=+-sqrt(25/4)$

Use the property

$color(brown)(sqrt(x/y)=sqrtx/sqrty$

$rarrx=+-sqrt25/sqrt4$

$color(blue)(rArrx=+-5/2$

Remember that the symbol $+-$ means Plus or Minus

it indicates that

$rArrcolor(blue)(x=5/2,-5/2$

How do you solve 4x^2=254x2=254x^2=25?