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Mathematics

Grade-11

Easy

Question

Solve the quadratic equation below using the Square Root Method.
$negative 7 open parentheses x squared minus 10 close parentheses squared minus 6 equals negative 258$

5, -5
6, -6
4, -4
2, -2

The correct answers are: 4, -4, 2, -2

$negative 7 open parentheses x squared minus 10 close parentheses squared minus 6 equals negative 258$
$negative 7 open parentheses x squared minus 10 close parentheses squared minus 6 plus 6 equals negative 258 plus 6$
$negative 7 open parentheses x squared minus 10 close parentheses squared equals negative 252$
$fraction numerator negative 7 open parentheses x squared minus 10 close parentheses squared over denominator negative 7 end fraction equals fraction numerator negative 252 over denominator negative 7 end fraction$
$open parentheses x squared minus 10 close parentheses squared equals 36$
$square root of open parentheses x squared minus 10 close parentheses squared end root equals plus-or-minus square root of 36$
$x squared minus 10 equals plus-or-minus 6$
Now we have to break up $x squared equals plus-or-minus 6 plus 10$ into two cases because of the "plus" or "minus" in 6.
Solve the first case where 6 is positive.
$x squared equals 6 plus 10$
$x squared equals 16$
$square root of x squared end root equals plus-or-minus square root of 16$
$x equals plus-or-minus 4$
Solve the second case where 6 is negative.
$x squared equals negative 6 plus 10$
$x squared equals 4$
$square root of x squared end root equals plus-or-minus square root of 4$
$x equals plus-or-minus 2$
The solutions to this quadratic equations are x = 4, x = -4, and x = 2, x = -2 . Yes, we have four values of that can satisfy the criginal quadratic equation.

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