Introduction to Quadratic Equations

The product of roots is (\frac{9}{p}=3), so (p=3). Use (\frac{c}{a}) for the product of roots.

Check each equation by substituting x = -2.
A: (-2)^2 + 2(-2) = 4 - 4 = 0 → x = -2 is a root.
B: (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 → x = -2 is a root.
C: 2(-2)^2 + 3(-2) - 2 = 8 - 6 - 2 = 0 → x = -2 is a root (this is the trickiest distractor because coefficients look nontrivial).
D: (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12 ≠ 0 → x = -2 is not a root.
Therefore option D is correct. Exam tip: For testing a single candidate root, direct substitution is quicker and less error-prone than factoring.

Factorise: \(x^2+2x-35=(x+7)(x-5)=0\). Hence the roots are \(x=-7\) and \(x=5\), i.e. the pair \((5,-7)\). Check: sum \(5+(-7)=-2=-b/a\) and product \(5\times(-7)=-35=c/a\). Option B \((7,-5)\) has the correct product but wrong sum (+2), so it is incorrect. Exam tip: verify roots quickly using sum and product of roots or factorisation before finalising the answer.

If the roots are 6 and 6, their sum is 6+6=12 and product is 6×6=36. For the quadratic \(x^2+rx+36=0\), the sum of roots equals \(-r\) and the product equals 36. Hence \(-r=12\) gives \(r=-12\). Option A (12) mistakes the sign. Options C and D are the root values themselves, not the coefficient asked for. Exam tip: Always apply sum = -b/a and product = c/a, and check signs carefully.

Expand and combine like terms: \((x-3)^2=x^2-6x+9\) and \(2(x-3)=2x-6\). Adding and subtracting 8 gives \(x^2-6x+9+2x-6-8=x^2-4x-5\). Therefore \(x^2-4x-5=0\) is correct. Closest distractor D has the same linear term but wrong constant sign (+5 instead of -5). Exam tip: Always expand \((x-a)^2\) as \(x^2-2ax+a^2\) and then combine like terms carefully.

First expand the left side: \((x+2)(x-7)=x^2-7x+2x-14=x^2-5x-14\). To put in standard form \(ax^2+bx+c=0\), subtract \(3x\) from both sides: \(x^2-5x-14-3x=0\Rightarrow x^2-8x-14=0\). Option B (\(x^2-5x-14=0\)) is the expansion before moving \(3x\), so it is the closest distractor. Options C and D result from sign or arithmetic mistakes. Exam tip: Expand first, then bring all terms to one side and order by descending powers of \(x\).

Substitute \(x=5\): the left-hand side becomes \(5^2-12\cdot5+35=25-60+35=0\). Thus the value is 0 and \(x=5\) is a root. Closest distractors are incorrect because they are either the substituted value (5) or give nonzero results when substituted; always verify by direct substitution and careful arithmetic. Exam tip: perform the arithmetic step-by-step to avoid sign mistakes.

For equal roots, (D=0), so (196-4k=0) and (k=49). For equal roots, set the discriminant to zero.

For equal roots, (k^2=196), and the equal root is (-\frac{k}{2}). For a positive root, (k=-14) is correct.

In (3x^2-27=0), the (x^2) term is present and the (x) term is absent. An equation can be quadratic even without the (x) term.

Dividing every term by (6) gives (x^2-3x+2=0). Dividing by a common nonzero factor does not change the roots.

We need two integers m and n such that m·n = 42 and m + n = 13. Checking 6 and 7 gives 6·7 = 42 and 6+7 = 13, so the quadratic factors as (x-6)(x-7), which expands to x^2 - 13x + 42. A close distractor is 3 and 14: their product is 42 but 3+14 = 17, not 13, so they do not produce the correct middle term. Exam tip: list factor pairs of the constant term and check which pair sums to the coefficient of x (taking sign into account).

For a quadratic \(ax^2+bx+c=0\), the sum of roots = \(-b/a\) and product = \(c/a\). Here \(a=1\) and \(b=s\), so the sum of roots is \(-s\). The given roots sum to \(-3)+(-6)=-9\), therefore \(-s=-9\) which gives \(s=9\). Common wrong answers: \(-9\) arises from forgetting the negative sign, \(18\) confuses sum with product (since product = 18), and \(-6\) mistakes a single root for the coefficient. Exam tip: identify \(a,b,c\) first and apply "sum = -b/a, product = c/a."

For a square with side x the area is \(x^2\). The statement says the area is 15 more than 8 times the side, so \(x^2=8x+15\). Bringing all terms to one side gives \(x^2-8x-15=0\), so option A is correct. Distractors fail for clear reasons: C swaps the coefficients (places 15 with 8) and thus misrepresents the given relation; B has wrong signs (+) so it does not match the sentence. Exam tip: Always translate the English sentence into a mathematical equality (area = ...) before rearranging to standard quadratic form.

Use the identity \((x+a)^2 = x^2 + 2ax + a^2\). Here the middle coefficient 12 gives \(2a=12\) so \(a=6\), and \(a^2=36\), hence \(x^2+12x+36=(x+6)^2\). Option B, \((x-6)^2\), expands to \(x^2-12x+36\) so its middle term has the opposite sign and is incorrect. Exam tip: half the middle coefficient and square it — if that equals the constant term and signs match, it's a perfect square trinomial.

Here ((2x-1)^2=4x^2-4x+1), and bringing all terms to one side gives (4x^2-7x-4=0). Apply the square identity and signs carefully.

Substitute \(x=-4\) into the equation: \((-4)^2 + t(-4) - 8 = 0\), which gives \(16 - 4t - 8 = 0\). Thus \(8 - 4t = 0\) and \(t = 2\). The common wrong choice \(-2\) typically arises from a sign error (writing \(+4t\) instead of \(-4t\)). Exam tip: directly substitute the root and carefully handle squares and signs when simplifying.

Expanding \((4x-1)(x+3)\) gives \(4x^2+12x-x-3\). Combining like terms yields \(4x^2+11x-3\), so the standard form is \(4x^2+11x-3=0\). Option B has the wrong sign for the middle term (should be +11x). Options C and D result from incorrect multiplication or sign errors. Exam tip: multiply each term carefully and then combine like terms to avoid sign mistakes.

For a quadratic equation, the coefficient of (x^2) must not be (0). Thus (5-p\neq0), so (p\neq5).

The coefficients are \(a=6,\; b=-13,\; c=5\). So \(2a+b-c=2\times6+(-13)-5=12-13-5=-6\). Option B (4) is a common mistake if the sign of \(c\) is handled incorrectly (adding instead of subtracting). Exam tip: always write down the values of \(a,b,c\) first and track plus/minus signs carefully when substituting.