Free TCE Mathematics Methods – Foundation Flashcard Review

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Make x the subject of y=(3x−2)/5.

Multiply by 5: 5y=3x−2. Add 2 and divide by 3, giving x=(5y+2)/3. Substitution into the original numerator gives 3[(5y+2)/3]−2=5y.

If C=2πr and C=15π cm, determine r.

Substitute first: 15π=2πr. Since π is nonzero, r=15/2=7.5 cm. The units remain centimetres because r is a length.

Rearrange v=u+at for t, and state the required restriction.

Subtract u to get v−u=at, then t=(v−u)/a. This rearrangement requires a≠0; when a=0 the original equation does not determine t.

A student turns A=(p+q)/r into q=A/r−p. Diagnose the error.

Clearing the denominator gives Ar=p+q, so q=Ar−p. The student divided A by r even though r already divides the entire numerator in the original formula.

For f(x)=2x²−3x+1, calculate f(−2).

Substitute −2 for every x: f(−2)=2(−2)²−3(−2)+1=8+6+1=15. Brackets protect the negative input when it is squared.

If g(t)=5−2t, solve g(a)=−7.

The equation is 5−2a=−7. Thus −2a=−12 and a=6. The notation g(a) is an output obtained from input a, not multiplication g×a.

What does f(3)=−4 say about the graph y=f(x)?

It says that input 3 has output −4, so the point (3,−4) lies on the graph. It does not identify an x-intercept because the output is not zero.

For h(x)=1/(x−4), why is h(4) undefined?

Substitution produces 1/(4−4)=1/0, and division by zero is undefined. Therefore 4 is excluded from the function’s domain.

Expand (2x−3)(x+5).

Distribute both terms: 2x²+10x−3x−15=2x²+7x−15. The middle coefficient comes from combining both cross-products.

Factorise x²−7x+12.

Find two numbers with product 12 and sum −7: −3 and −4. Hence x²−7x+12=(x−3)(x−4).

Factorise 6x²+x−2.

Split the middle term: 6x²+4x−3x−2=2x(3x+2)−1(3x+2)=(2x−1)(3x+2). Expanding checks the coefficient of x is 1.

How can you check a proposed quadratic factorisation without guessing roots again?

Expand the proposed factors and compare the x², x and constant coefficients with the original polynomial. Exact agreement of all three coefficients verifies the identity.

Factorise 9x²−25.

This is (3x)²−5², so 9x²−25=(3x−5)(3x+5). The conjugate middle terms cancel when expanded.

Expand (x−6)².

Use (a−b)²=a²−2ab+b²: (x−6)²=x²−12x+36. It is not x²+36 because the cross-term is essential.

Complete the identity: 4x²+12x+9=(__).

The expression is (2x+3)² because the cross-term is 2(2x)(3)=12x. Both the leading and constant terms are perfect squares.

Why does x²+16 not factor as (x−4)(x+4)?

The product (x−4)(x+4)=x²−16 is a difference of squares. The sum x²+16 has no real linear factorisation of that form.

Write x²+8x−3 in completed-square form.

Half of 8 is 4: x²+8x−3=(x+4)²−16−3=(x+4)²−19. Expanding returns the original expression.

Complete the square for 2x²−12x+5.

Factor 2 from the quadratic terms: 2(x²−6x)+5=2[(x−3)²−9]+5=2(x−3)²−13.

Use completing the square to identify the minimum of 3x²+6x+7.

3x²+6x+7=3(x+1)²+4. Since the squared term is non-negative, the minimum value is 4, attained at x=−1.

Solve 2x²−3x−2=0 using the quadratic formula.

With a=2,b=−3,c=−2, x=[3±√(9+16)]/4=[3±5]/4. Thus x=2 or x=−1/2.