Given that $\alpha, \beta$ are the roots of the equation $3x^2+2x+7=0$, find the equation of the quadratic equation with roots $\alpha^2, \beta^2$.

 Given that $\alpha, \beta$ are the roots of the equation $3x^2+2x+7=0$, find the equation of the quadratic equation with roots $\alpha^2, \beta^2$.

Algebra type questions from previous exams (BOSTES 2U)

QUESTION(Quadratic Equations) Solve the equation $$x^2+12x+36=64$$

QUESTION(Quadratic Equations)
.Solve the equation $$x^2+12x+36=64$$

SOLUTION: =================
I shall do this with two methods as I am not sure which you are familiar with. 

We have 

$x^2 + 12x + 36 = 64 $
$x^2 + 12x + 36 - 64 = 0 $
$x^2 + 12x - 28 = 0 $

Method 1 Factorise to get   
$(x - 2)(x + 14)=0 $

so, 
$x - 2 = 0$ and  $x + 14 = 0 $
This gives $x = 2, -14 $

Method 2 : Use the quadratic formula (see below in appendix) 
This gives 
 $ x = [ - 12 \pm \sqrt{12^2 - 4(1)(-28)} ] / (2(1)) $
 $x = [ -12 \pm \sqrt{256} ] / 2 $
$x = [ -12 \pm 16 ]/2 $
$x = [-12 + 16]/2  ,\quad[-12-16]/2 $
$x = 2, -14   $  

CHECK:  
when $x = 2$,
 LHS$ = x^2 + 12x + 36 $ 
$=2^2 +12(2) + 36 $
$= 4 + 24 + 36 $
$= 28 + 36 $
$= 64 =$ RHS 
when $x = -14$, 
LHS $= x^2 + 12x + 36 $
$= (-14)^2 +12(-14) + 36 $
$= 196 - 168 + 36 $
$= 28 + 36$
$ = 64 = $RHS 


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Appendix: The Quadratic Formula 

The quadratic equation $ax^2 + bx + c = 0$  has solutions given by 
                                  $$x ={ -b \pm \sqrt{b^2 - 4ac} \over 2a} $$

Real solutions only exist provided $b^2 - 4ac \ge 0$.

QUESTION(Quadratic Equations) Solve the equation $$(x+7)^2=64$$

QUESTION(Quadratic Equations)
Solve the equation
$$(x+7)^2=64$$

SOLUTION: ===============
Starting with   $(x+7)^2=64$ 
Take square roots of both sides to give 
$x+7 =  \pm \sqrt{64} $
$x+7 =  \pm 8 $
$x=  -7 \pm 8 $
$x = -7-8  or -7 + 8 $
$x = -15, 1 $

[  CHECK: 
when $x=-15$, 
LHS $= (x+7)^2 = (-15 + 7)^2 = (-8)^2 = 64 = $ RHS.
when $x = 1$, 
LHS $= (x+7)^2 = (1 + 7)^2 = (8)^2 = 64 =$ RHS.  ]