Outside the high school program / Algebra: second-degree expression with discriminant 66 exercises (including 56 corrected)

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<0Aucune solution01solution−b2·a>02solutions−b−2·a;−b2·a f(xg(xPour cette valeur (abscisse),les deux fonctions aurontla même image (ordonnée) -123I-2-12JOCfCg 1. Second degree: equation E.4411 Proposition: The roots of a polynomial are the values that cancel out this polynomial. For a second-degree polynomial a · x 2 + b · x + c , the number of existing roots depends on the discriminant : Determine the roots of second-degree polynomials: a x 2 + 2 x − 35 = 0 b 2 x 2 − 5 x + 2 = 0 c 5 x 2 − 3 x + 2 = 0 d 9 x 2 − 24 x + 16 = 0 e − 2 x 2 + 3 x − 5 = 0 f − 3 x 2 − x + 4 = 0 E.4415 Solve the following equations : a 3 x 2 + 4 x + 1 = 0 b 3 x 2 − 4 x + 2 = 0 c − x 2 + 2 x + 3 = 0 d 2 x 2 − 4 x + 2 = 0 E.4412 Solve the following equations : a 2 x 2 − 3 x − 2 = 0 b x 2 − 3 x + 1 = 0 E.4479 Reminders : For a quadratic function f : x ↦→ a · x 2 + b · x + c , the an-tecedents of 0 are the roots of the polynomial. The x-coordinates of the intersection points of the curves f and g satisfy the equation f ( x )= g ( x ) . Be-cause : Consider the two functions f and g whose images of a number x are defined by the relations : f ( x ) = − x 2 + 3 x − 2 ; g ( x ) = 3 x 2 − x − 1 In the coordinate system O ; I ; J , we give the graphical rep-resentations C f and C g of the functions f and g , respectively: 1 Graphically, give the antecedents of the number 0 by the function f . The following questions will be answered algebraically: 2 Determine the antecedents of the number 0 using the function g . Give the values rounded to two decimal places. 3 a Solve the equation : f ( x )= g ( x ) b Determine the coordinates of the points of intersection of the curves C f and C g . E.4482 Solve the following equations : a − 5 x 2 − 4 x + 1 = 0 b − 6 x 2 + 8 x + 8 = 0 c (3 x − 2)(2 x − 2) = x (2 x + 2) − 5 E.7486 Consider the two functions f and g de-fined on − 1 ; + ∞ defined by the relations : f ( x ) = − 2 x + 1 ; g ( x ) = 2 · x − 3 In a reference frame O ; I ; J , we note C f and C g the repre-sentative curves of the functions f and g . 1 Establish the identity: g ( x ) − f ( x )= 2 · x 2 − x − 1 x +1 2 Deduce the abscissas of the intersection points of the curves C f and C g . E.8107 Solve the equation : − 6 · x 2 − 8 · x − 2=0 2. Second degree: equation and radical simplification E.9440 Solve the following equations : a − 3 x 2 + 3 x + 3 = 0 fb − x 2 + 4 x + 3 = 0 E.9441 Solve the following equations : a x 2 − 4 x − 3 = 0 b 2 x 2 − 4 x − 8 = 0 3. Second degree: factorized form https://chingmath.fr chapExoCorrec/4411 sacados/4411 <0Aucune solution01solution−b2·a>02solutions−b−2·a;−b2·a chapExoCorrec/4415 sacados/4415 chapExoCorrec/4412 sacados/4412 chapExoCorrec/4479 sacados/4479 f(xg(xPour cette valeur (abscisse),les deux fonctions aurontla même image (ordonnée) -123I-2-12JOCfCg chapExoCorrec/4482 sacados/4482 chapExoCorrec/7486 sacados/7486 chapExoCorrec/8107 sacados/8107 chapExoCorrec/9440 sacados/9440 chapExoCorrec/9441 sacados/9441
<0Aucunefactorisation0a·xb2·a2>0a·x−¸x−˛où¸et˛sontles deux racines du polynômes <00>0¸et˛sontlesdeuxracinesa>0a<0x−∞∞x−∞∞−x−∞∞−b/2a0x−∞∞−b/2a0−−x−∞∞αβ00−x−∞∞αβ00−− E.4466 Factor, if possible, the following expres-sions : a 2 x 2 + 3 x + 1 b 4 x 2 + 9 x + 2 c − 3 x 2 + 2 x − 1 Hint: present results as : a · x + b c · x + d ou a · x + b 2 with a;b;c;d ∈ Z E.4413 Proposition: The factorization of a second-degree polyno-mial a · x 2 + b · x + c depends on the value of its discriminant : Factorize, if possible, the following expressions : a x 2 + 4 x + 3 b 5 x 2 − 4 x − 1 c 3 x 2 + 4 x + 1 d 4 x 2 + 3 x + 4 e 12 x 2 + 36 x + 27 f 3 x 2 + 3 x + 4 E.4416 Factor, if possible, the second-degree polynomials below : a 3 x 2 + 4 x + 1 b − 4 x 2 + 5 x c − 4 x 2 + 12 x − 9 Hint: present results as : a · x + b c · x + d ou a · x + b 2 with a;b;c;d ∈ Z 4. Second degree: inequation E.4417 Proposal: The sign table for a quadratic polynomial de-pends on the sign of the coefficient of the quadratic term and the sign of the discriminant. The six possibilities are shown below : Draw the sign chart for each of the following expressions : a 2 x 2 + x − 1 b − 3 x 2 + 2 x + 1 c 2 x 2 − 1 d 4 x 2 − 3 x − 1 e − x 2 − x − 1 f − x 2 − 4 x − 1 Note: For question f , , use the roots of the polynomial, rounded to two decimal places. E.4467 Solve the following inequations : a x 2 + 5 x + 4 > 0 b − 3 x 2 + 4 x + 4 < 0 c 4 x 2 + 4 x + 1 > 0 d − 2 x 2 + 5 x + 3 > 0 e 4 x 2 − 3 x + 2 0 f 12 x 2 + 12 x + 3 0 E.4611 Draw up the sign table for the following expressions at R : a 2 x 2 − 3 x − 2 b (2 x + 1)(3 x 2 − 2 x − 1) E.4483 Solve the following inequations : a 2 x 2 + 3 x − 5 > 0 b − x 2 + 5 x − 4 0 c ( − 4 x 2 + x + 5)(3 − 2 x ) 0 5. Second degree: table of variations and sign table E.4498 Consider the functions f and g defined on R defined by the relations : f ( x ) = x 2 + x + 1 ; g ( x ) = − 2 · x 2 − 3 · x + 5 1 Draw up the table of variations for each of these func-tions. 2 Establish the sign table for each of these functions. E.4500 1 Consider the function f whose image of a real number x is defined by: f ( x ) = 2 · x 2 + 3 · x + 2 a Draw up the table of variations of the function f . b Justify that the function f admits no antecedent of the number 0 . 2 Let g be the function defined by the relation: g ( x ) = − x 2 + 2 · x + 3 a Determine the set of antecedents of the number 0 by the function g . b Draw up the table of variations of the function g (the results of the previous question will be brought into this) . c Deduce the sign table of the function g . https://chingmath.fr chapExoCorrec/4466 sacados/4466 chapExoCorrec/4413 sacados/4413 <0Aucunefactorisation0a·xb2·a2>0a·x−¸x−˛où¸et˛sontles deux racines du polynômes chapExoCorrec/4416 sacados/4416 chapExoCorrec/4417 sacados/4417 <00>0¸et˛sontlesdeuxracinesa>0a<0x−∞∞x−∞∞−x−∞∞−b/2a0x−∞∞−b/2a0−−x−∞∞αβ00−x−∞∞αβ00−− chapExoCorrec/4467 sacados/4467 chapExoCorrec/4611 sacados/4611 chapExoCorrec/4483 sacados/4483 chapExoCorrec/4498 sacados/4498 chapExoCorrec/4500 sacados/4500
-2-1234I-2-12JOCfCg 6. Second degree: relative position of curves E.4471 In the plane provided with a reference frame O ; I ; J , consider the curves C f and C g representa-tive of the functions f and g defined by: f ( x ) = − x 2 + 4 · x − 3 ; g ( x ) = 7 2 · x 2 − 5 x + 1 The questions below must be answered algebraically: 1 Determine the antecedents of 0 by the functions f and g . Exact values rounded to the nearest hundredth will be given. 2 Determine the relative position of these two curves. E.7417 Consider the two functions defined re-spectively on R and R \{− 1 } and by the relations : f ( x ) = 2 2 · x + 3 ; g ( x ) = 2 · x + 2 1 a Establish the equality: g ( x ) − f ( x )= 4 · x 2 +10 · x +4 2 · x +3 b Draw up the sign table for the polynomial 4 · x 2 +10 · x + 4 leaving the steps of your reasoning. 2 In the plane provided with a reference frame O ; I ; J , note C f and C g the representative curves of the functions f and g . Compare the curves C f and C g at − 2 3 ; + ∞ . 7. Equation and solution set E.2735 Consider the following equation : 1 2 x 2 − 3 x + 1 − 1 2 x 2 + 5 x + 2 = 0 1 Determine the solution set for this equation. 2 Determine the solution set of this equation. E.2733 Solve the following equations, paying attention to the solution set for each equation : a 3 x 2 − 2 x − 3 = x b 1 3 x 2 − 8 x + 4 = − 2 5 x 2 − 6 x − 8 E.5055 Consider the function f whose image of a real number x is defined by the relation: f ( x ) = − 2 x 2 − 3 x + 2 1 a Determine the definition set of the function f . b Draw up the table of variations of the function f . 2 Consider the straight line ( d ) of equation : y = − 5 · x . a Solve equation : − 2 x 2 − 3 x +2=25 · x 2 . b Justify that the line ( d ) intercepts the curve C f repre-sentative of the function f at a single point. E.9439 Solve the inequation : − 4 · x 2 +3 · x +7 0 8. Inequations and rational expressions E.2278 Solve the following inequalities: a x 2 − x 2 x − 1 0 b 18 x 2 − 12 x + 2 3 x 2 + x − 2 > 0 c x 2 + 2 x − 5 x 0 d x − 2 x + 1 − 3 x − 1 x − 1 < 0 E.2298 Solve the following inequalities: a 3 x 2 − 5 x + 2 − 3 x 2 + 4 x − 2 0 b 2 x − 5 2 x − 1 < x + 1 x + 3 E.2306 Solve the following inequation : 3 x 2 − 5 x + 2 − 3 x 2 − 4 x + 2 0 E.2747 Solve in R the inequation : − 3 x 2 + 4 x + 4 5 x 2 + x − 4 0 E.835 Solve the following equation in R : − x 2 − 3 3 + 2 = 1 2 x + 1 https://chingmath.fr chapExoCorrec/4471 sacados/4471 -2-1234I-2-12JOCfCg chapExoCorrec/7417 sacados/7417 chapExoCorrec/2735 sacados/2735 chapExoCorrec/2733 sacados/2733 chapExoCorrec/5055 sacados/5055 chapExoCorrec/9439 sacados/9439 chapExoCorrec/2278 sacados/2278 chapExoCorrec/2298 sacados/2298 chapExoCorrec/2306 sacados/2306 chapExoCorrec/2747 sacados/2747 chapExoCorrec/835 sacados/835
-5-4-3-2-1012345-3-2-1123C1 -5-4-3-2-1012345-3-2-1123C2 -5-4-3-2-1012345-3-2-1123C3 -5-4-3-2-1012345-3-2-1123C4 E.5743 Consider the function f whose image of a real number x is given by the relation: f ( x ) = x + 2 x 2 + 2 x − 3 1 Solve the inequation : f ( x ) 0 2 Of the four curves below, only one is the representative curve of the function f . Which curve? E.2283 Give the definition set of the function f defined by: f ( x ) = 1 − 2 x + 1 x 2 + 5 x − 1 E.6632 Consider the two functions f and g de-fined respectively on R \{− 1 } and R by the relations : f ( x ) = 2 · x 2 + 3 · x − 3 x + 1 ; g ( x ) = x − 2 Solve the inequation : f ( x ) g ( x ) E.9438 Solve in R the following equation and inequation : a − x 2 − 3 3 + 2 = 1 2 x + 1 b 4 x x + 1 5 x − 3 9. Inequations and rational expressions: degree greater than 2 E.4569 Consider the two functions f and g defined on − 2 ; + ∞ by the relations : f ( x ) = x − 1 x 2 + 1 ; g ( x ) = 3 · x − 3 2 · x + 4 1 a Establish the following equality: f ( x ) − g ( x ) = − 3 · x 3 + 5 · x 2 − x − 1 ( x 2 + 1)(2 · x + 4) b Determine the value of the reals a , b and c verifying the following relationship : f ( x ) − g ( x ) = x − 1 a · x 2 + b · x + c ( x 2 + 1)(2 x + 4) 2 Determine the set of solutions to the equation : f ( x ) − g ( x ) = 0 3 a Draw up the sign table for : f ( x ) − g ( x ) . (we’ll admit that the product ( x 2 +1)(2 · x +4) is strictly positive at − 2 ; + ∞ ) . b Deduce the relative position of the curves C f and C g on the interval − 2 ; + ∞ . E.4541 Consider the two functions f and g de-fined on R by the relations : f ( x ) = 2 · x 3 − 10 · x 2 + 6 · x + 12 ; g ( x ) = 2 · x − 4 Consider C f and C g the representative curves of the functions f and g . 1 Show that 2 is a solution of the equation : f ( x )= g ( x ) . 2 a Determine the value of the real numbers a , b , c ver-ifying: f ( x ) − g ( x ) = x − 2 a · x 2 + b · x + c b Deduce the factorized form of the expression : f ( x ) − g ( x ) . 3 a Draw up the sign table for the difference : f ( x ) − g ( x ) b Deduce the relative position of the curves C f and C g on R . https://chingmath.fr chapExoCorrec/5743 sacados/5743 -5-4-3-2-1012345-3-2-1123C1 -5-4-3-2-1012345-3-2-1123C2 -5-4-3-2-1012345-3-2-1123C3 -5-4-3-2-1012345-3-2-1123C4 chapExoCorrec/2283 sacados/2283 chapExoCorrec/6632 sacados/6632 chapExoCorrec/9438 sacados/9438 chapExoCorrec/4569 sacados/4569 chapExoCorrec/4541 sacados/4541
E.2736 Consider the two functions f and g whose images of a number x are defined by the following re-lationships : f ( x ) = 1 x − 2 ; g ( x ) = 2 x + 1 1 a Determine the table of variations of the functions f and g respectively on the intervals 2 ; + ∞ et − 1 2 ; + ∞ . b Justify, in view of the table of variation, that the curves C f and C g intersect only once on the interval ]2 ; + ∞ [ . 2 We wish to solve the equation : ( E ): 1 x − 2 = 2 x +1 a Determine the values of a , b , c so that : ( x − 2) 2 · (2 x + 1) − 1 = (2 x − 3)( a · x 2 + b · x + c ) b Deduce the solutions of equation : ( x − 2) 2 · (2 x + 1) − 1 = 0 . c Determine the solutions of the equation ( E ) . E.4546 Consider the function f whose im- age of a number x is defined by the relation: f ( x ) = − x − 3 2 · x 2 − 3 1 Determine the definition set of the function f . 2 Let g be the affine function defined by the relation: g ( x ) = x + 3 a Determine the value of the reals a , b and c verifying the following equality: f ( x ) − g ( x )= (1 − x )( a · x 2 + b · x + c ) 2 · x 2 − 3 b Deduce the factorized form of f ( x ) − g ( x ) . 3 a Draw up the sign table for f ( x ) − g ( x ) on the interval − 3 2 ; 3 2 . We’ll admit that the expression 2 x 2 − 3 is negative on − 3 2 ; 3 2 . b Deduce the relative position of the curves C f and C g on the interval − 3 2 ; 3 2 . 10. A little further on E.2739 The plane is given an orthonormal refer-ence frame ( O ; I ; J ) and the points A ( − 0.6 ; 1.2) and B (1 ; 0) are considered 1 Determine the measure of distance AB . 2 We wish to determine the coordinates of the point C such that the triangle ABC is equilateral: a Justify that the coordinates of the point C verify the following two equations : x C + 0.6 2 + y C − 1.2 2 = 4 x C − 1 2 + y C 2 = 4 b Deduce the following equality: y = 4 3 x + 1 3 c Deduce the possible values of x . d Give the coordinates of the two points realizing the statement. E.2924 The purpose of this exercise is to demon-strate that the only periodic polynomials are constant poly-nomials. We will use the following proposition : Proposition: Any real polynomial of degree n has at most n real roots. Let a be a nonzero real number and P a periodic polynomial of period a : 1 Let Q be the polynomial defined by: Q ( x ) = P ( x ) − P ( a ) Justify that, for all n ∈ N , we have : Q ( n · a ) = 0 2 Conclude that the polynomial P is constant. 11. A little further: systems of equations of the second degree E.2715 Consider the following system : x · y = − 1 x − 2 2 y + 3 = 0 1 Check that the couple 2 3 ; − 3 2 is solution of this sys-tem 2 Justify that the abscissas of the solutions of this system form the set of solutions of the following equation on R ∗ : 3 x 2 − 8 x + 4 3 Deduce the set of solutions of this system. https://chingmath.fr chapExoCorrec/2736 sacados/2736 chapExoCorrec/4546 sacados/4546 chapExoCorrec/2739 sacados/2739 chapExoCorrec/2924 sacados/2924 chapExoCorrec/2715 sacados/2715
IJOCgCgCfCf E.2716 1 Determine the solutions of the following system of equa-tions : x · y = − 2 2 y + 3 x − 2 = 1 2 Consider the two functions g defined whose images of a number x are defined by the following relationships : f ( x ) = 7 − 3 x 2 x − 4 ; g ( x ) = − 2 x Here is the representative curve of these two functions : a Justify that the points of intersection of these two curves verify the system of equations in question 1 . b Deduce the coordinates of the intersection points of the curves C f and C g . E.2748 Solve the system of equations : 2 x + 3 y = 12 x × y = − 2 E.1393 Solve the following system : x 2 + 4 y 2 = 5 x × y = 1 Indication : This system has four pairs of solutions. E.2003 Solve the following system : − 5 x 2 + 4 y 2 = − 1 x × y = 1 Hint: this system admits two couples as solutions. E.2004 Solve the following system : − 2 x 2 − 2 y 2 = − 4 x × y = − 1 Hint: this system admits two couples for solutions E.2005 Solve the following system : 4 x 2 + y 2 = − 3 x × y = 1 Hint: this system admits no couple as solution. E.2006 Solve the following system : − x 2 + 9 y 2 = 9 x × y = − 2 E.2007 a − 10 x 2 + 4 y 2 = 6 x × y = − 2 b − 7 x 2 + 4 y 2 = 9 x × y = − 2 c − 4 x 2 + 3 y 2 = 8 x × y = − 2 d − 2 x 2 + 1 y 2 = − 1 x × y = − 1 e 1 x 2 + − 8 y 2 = − 7 x × y = 1 f 2 x 2 + − 7 y 2 = 1 x × y = 2 g 3 x 2 + − 5 y 2 = 7 x × y = − 2 h 4 x 2 + − 4 y 2 = − 7 x × y = 3 i 5 x 2 + − 4 y 2 = 4 x × y = − 4 j 6 x 2 + − 5 y 2 = 9 x × y = − 9 E.2008 Solve the following system : 6 x 2 + 7 y 2 = − 8 x × y = − 5 E.7248 Solve the system of equations : x 2 + y 2 + z 2 = 1 x + y + z = 1 E.2044 Solve the following system : x × y = − 2 ( x + 8)( y + − 2) = − 7 E.2042 Solve the following systems : a x × y = 3 ( x + − 1)( y + 1) = 4 b x × y = 3 ( x + 5)( y + 5) = 8 c x × y = 4 ( x + 6)( y + 3) = 4 d x × y = 6 ( x + − 3)( y + 1) = 6 e x × y = 7 ( x + 2)( y + 2) = − 5 E.2043 Solve the following system : x × y = 4 ( x + 6)( y + − 3) = − 8 E.2045 Solve the following system : x × y = 2 ( x − 1)( y − 9) = 2 E.2046 Solve the following equation : x × y = 2 ( x + 4)( y − 1) = − 2 12. A little further: changing variables E.6793 1 Determine the roots of the polynomial: ( P ) : x 2 − 8 · x + 4 2 Expand and simplify the following expressions : a = 1 + 3 2 ; b = 1 − 3 2 3 Consider the polynomial ( P ) defined by: https://chingmath.fr chapExoCorrec/2716 sacados/2716 IJOCgCgCfCf chapExoCorrec/2748 sacados/2748 chapExoCorrec/1393 sacados/1393 chapExoCorrec/2003 sacados/2003 chapExoCorrec/2004 sacados/2004 chapExoCorrec/2005 sacados/2005 sacados/2006 sacados/2007 sacados/2008 sacados/7248 sacados/2044 Aucun couple de solution sacados/2042 sacados/2043 Un seul couple de solution sacados/2045 Couple de solution avec valeurs rationnelles sacados/2046 Solution sous forme de radicaux chapExoCorrec/6793 sacados/6793
ABCDEFGx4m -2-12I-4-22JOCfASB ( P ) : x 4 − 8 · x 2 + 4 a Show that 1+ 3 is a root of ( P ) . b Deduce the four roots of the polynomial ( P ) . E.6877 Solve the following two equations using the change of variable: a P : x 4 − 5 · x 2 + 4 b P : x 4 + 5 · x 2 + 4 E.2971 Consider the expression ( E ) de-fined by: ( E ) : x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 1 a Show that if a is a solution of the equation ( E ) then so is 1 a . b Show that the equation ( E ) is equivalent to the equa-tion : ( E ) : x 2 − 3 x + 4 − 3 x + 1 x 2 = 0 2 a Expand the following expression : x + 1 x − 1 x + 1 x − 2 b Using the change of variable X = x + 1 x , modify the equation ( E ) into a second-degree equation in X . c Solve the equation in X obtained in the previous ques-tion. d Deduce the values of x solution of ( E ) . 3 Give the set of solutions to the equation ( E ) . E.5143 Consider the equation ( E ) defined by: x 4 − 8 x 3 + 2 x 2 − 8 x + 1 = 0 1 Show that the equation ( E ) is equivalent to : ( E ): x 2 − 8 x + 2 − 8 x + 1 x 2 = 0 2 Determine the reals a , b and c verifying the relation: x 2 − 8 x + 2 − 8 x + 1 x 2 = a · x + 1 x 2 + b · x + 1 x + c 3 Posing for variable change X = x + 1 x , solve equation ( E ) . 13. Second degree: problems E.4423 A field is made up of two squares and a right-angled triangle. This field is shown in the figure below : 1 Justify that the area A of the field has the value as a function of x : A ( x ) = x 2 + 2 x + 32 2 Deduce the value of length x so that the total area of the field is 200 m 2 . E.4499 The curve C f is a parabola representing a f function of the second degree. The curve C f passes through the points A ( − 1 ; − 2) , B (0 ; − 1) and has as its vertex the point S whose abscissa is − 1 4 . Since the function f is defined by a polynomial of the second degree, we deduce the existence of three real numbers a , b and c such that : f ( x ) = a · x 2 + b · x + c 1 Using the coordinates of the point B , determine the value of the number c . 2 Using the characteristics of the vertex S of the parabola, justify that the function f admits the writing: f ( x ) = 2 b · x 2 + bx − 1 3 Using the coordinates of point A , determine the complete expression of function f . https://chingmath.fr chapExoCorrec/6877 sacados/6877 chapExoCorrec/2971 sacados/2971 A refaire pique sur internet chapExoCorrec/5143 sacados/5143 chapExoCorrec/4423 sacados/4423 ABCDEFGx4m chapExoCorrec/4499 sacados/4499 -2-12I-4-22JOCfASB
16cmxx10cm E.4470 A rectangular box without a lid is to be made in the pattern below. The lengths are expressed in cm . 1 a When the box is built, the number x will represent which dimension? Length, width or height? b What values can the variable x take in this problem? c Give the expression for the volume V as a function of the value of x . 2 In this question, we investigate for what values of ˇ x ı, this box has a volume equal to 144 cm 3 : a Establish the following equality: 4 x 3 − 52 x 2 + 160 x − 144 = (2 x − 4)(2 x 2 − 22 x + 36) b Deduce the values of x for which V ( x ) has the value 144. 14. Unclassified financial years E.4567 Let f be a function defined on R using a second-degree polynomial whose coefficient of the second-degree term is strictly negative. Its representative curve C f is a parabola whose vertex has coordinates 1 4 ; − 7 8 . 1 a Draw up a table of variations for the function f . b Deduce the sign of the discriminant of the quadratic polynomial defining the function f . 2 The curve C f passes through the points : A (0 ; − 1) et B ( − 2 ; − 11) . Determine the real numbers a , b and c that satisfy the equation : f ( x ) = a · x 2 + b · x + c E.4414 Consider the second-degree polynomial ( E ): x 2 +3 x +3 . 1 Determine the discriminant of the polynomial ( E ) . 2 By reasoning by the absurd and assuming that the ex-pression ( E ) admits the factorized form : ( E ) : a ( x − ¸ )( x − ˛ ) Establish that the expression ( E ) admits no factorized form. E.2806 Solve the following equation : − 3 √ − 2 x + 1 = − x − 3 E.2010 1 Establish the following identity: ax 2 + bx + c = a x + b 2 a 2 − b 2 − 4 ac 4 a 2 This question makes it possible to affirm that any polynomial of the second degree a · x 2 + b · x + c admits an entry of the form [( x + ˛ ) 2 − ‚ . The latter form is called the canonical form of a second-degree polynomial. 2 a Justify that the expression : a x + b 2 a 2 − b 2 − 4 ac 4 a 2 = 0 admits no solution, one solution, two solutions depend-ing on the value of b 2 − 4 ac ? b For each of the equations below, describe the set of solutions : i.e. whether it is empty or the number of elements making it up : ( E ) : 4 x 2 − 5 x + 4 ( F ) : 2 x 2 − x − 1 ( G ) : 9 x 2 − 24 x + 16 The set of solutions to the equation ax 2 + bx + c =0 depends on the sign of b 2 − 4 ac . We call the discriminant of the polynomial ax 2 + bx + c the number Δ worth: Δ = b 2 − 4 a · c . Relative to the previous question, we deduce that a second-degree equation admits as solution set: the empty set if Δ < 0 , a one-element set if Δ=0 , a set with two distinct elements if Δ > 0 . 3 Consider a second-degree polynomial with a strictly pos-itive discriminant : Δ= b 2 − 4 a · c> 0 a Factor the expression : x + b 2 a 2 − Δ b Deduce that the second-degree equation : a · x 2 + b · x + c = 0 admits as solutions : x 1 = − b − √ Δ 2 · a ; x 2 = − b + √ Δ 2 · a 4 Consider the equation ( H ): 2 x 2 +2 x − 12=0 a Determine the discriminant of the polynomial 2 x 2 + 2 x − 12 . How many solutions does the equation ( H ) admit? b By direct application of question 3 b , prove that the equation ( H ) admits as solution set − 3 ; 2 https://chingmath.fr chapExoCorrec/4470 sacados/4470 16cmxx10cm chapExoCorrec/4567 sacados/4567 chapExoCorrec/4414 sacados/4414 chapExoCorrec/2806 sacados/2806 sacados/2010