Introduction to Quadratic Equations

In the standard form \(ax^2+bx+c=0\), \(a\) is the coefficient of \(x^2\). In the given equation the coefficient of \(x^2\) is 7, so \(a=7\). Option B (2) is the coefficient of \(x\) (that's \(b\)), and option C (−5) is the constant term \(c\). Exam tip: Always write the quadratic in standard form \(ax^2+bx+c=0\) and read off \(a,b,c\) directly.

A quadratic is written as \(ax^2+bx+c=0\), where \(b\) is the coefficient of \(x\). In \(5x^2-9x+4=0\) the coefficient of \(x\) is \(-9\), so \(b=-9\). The option 5 is the coefficient \(a\), 4 is the constant term \(c\), and 9 is just the positive opposite of the correct value. Exam tip: rewrite the equation in the form \(ax^2+bx+c\) and read off the coefficient of \(x\) to avoid sign mistakes.

A quadratic equation is written in standard form as \(ax^2+bx+c=0\). Comparing \(8x^2+3x+12=0\) with this form gives \(a=8\), \(b=3\) and \(c=12\). Hence \(c=12\). Option 8 is incorrect because it is the coefficient of \(x^2\) (\(a\)), option 3 is the coefficient of \(x\) (\(b\)), and 0 is incorrect because the constant term here is 12, not 0. Exam tip: rewrite any given equation in standard form first and then read off \(a, b, c\).

A quadratic is written as \(ax^2+bx+c=0\). The given equation can be written as \(11x^2-6x+0=0\), so the constant term is \(c=0\). Note that \(-6\) is the coefficient \(b\) and \(11\) is \(a\); option D has the wrong sign. Exam tip: always put the equation in standard form and if the constant term is missing treat it as 0.

Write the equation in standard form as \(13x^2+0x-20=0\). Thus \(a=13,\; b=0,\; c=-20\), so \(b=0\). A common distractor is \(-20\), but that is the constant term \(c\), not the linear coefficient. Exam tip: if the x-term is missing, take the linear coefficient \(b\) equal to zero immediately.

By definition a quadratic equation must have the highest power of the variable equal to 2. That requires the coefficient of \(x^2\), namely \(p\), to be nonzero, i.e. \(p\neq0\). If \(p=0\) the \(x^2\) term disappears and the equation becomes linear (or lower degree), so option A is incorrect. Options C and D (\(q=0\) or \(r=0\)) merely eliminate other terms but do not change the degree as long as \(p\neq0\). Exam tip: always check the leading coefficient to determine if an equation is quadratic.

A quadratic is written in standard form as \(ax^2+bx+c=0\). Moving the RHS term \(9\) to the left changes its sign, giving \(4x^2+7x-9=0\), so option B is correct. Option A is wrong because it keeps \(+9\) instead of \(-9\). Options C and D are wrong because they change the sign of the linear term \(7x\). Exam tip: bring every term to one side and combine like terms to reach \(ax^2+bx+c=0\).

The standard form for a quadratic is \(ax^2+bx+c=0\). Move the right-hand side terms to the left by subtracting \(5x+3\) from both sides to get \(2x^2-5x-3=0\). Thus \(a=2, b=-5, c=-3\). Option C is the closest distractor but has the constant term with the wrong sign (+3 instead of −3). Exam tip: always convert to \(ax^2+bx+c=0\) by bringing all terms to one side before identifying coefficients or solving.

Substitute \(x=3\): \(3^2-4\cdot3+3 = 9-12+3 = 0\). Thus the left-hand side equals \(0\), so \(x=3\) makes the expression zero. The closest distractor \(3\) typically arises from sign or squaring mistakes (e.g. treating \(3^2\) as \(3\)). Exam tip: evaluate each term separately and follow the order of operations (BODMAS).

Solving \(x^2-25=0\) gives \(x^2=25\), so \(x=\pm5\). The value 5 is among the choices and indeed satisfies \(5^2-25=0\), so 5 is a correct solution. Checking the others: \(2^2-25=-21\), \((-2)^2-25=-21\), \(10^2-25=75\) — none are zero, so they are not solutions. Exam tip: take square roots to get \(x=\pm\sqrt{25}\) and then substitute the available options to verify quickly.

Substituting \(x=4\) gives the left-hand side \(4^2-8\cdot4+16=16-32+16=0\), so the equation becomes \(0=0\); therefore \(x=4\) satisfies the equation. Option B (4=0) is a tempting distractor but incorrect because the left-hand side evaluates to 0, not 4. Exam tip: when checking roots, simplify squares and products first, then perform addition/subtraction to avoid arithmetic mistakes.

The general form of a quadratic is \(ax^2+bx+c=0\). The equation \(12x^2=0\) can be written as \(12x^2+0x+0=0\), so the coefficients are \(a=12,\ b=0,\ c=0\). The distractor (12,1,0) is wrong because the coefficient of x is 0, not 1. Exam tip: always write missing terms as \(0x\) or \(+0\) to read off coefficients correctly.

Core idea: use the distributive property. Multiply \(2x\) across the bracket: \(2x\cdot x=2x^2\) and \(2x\cdot7=14x\), giving \(2x^2+14x=0\). The closest distractor \(2x^2-14x=0\) is wrong because the sign of the linear term is incorrect. Option \(14x^2+2x=0\) swaps coefficients and is not equivalent; \(2x+7=0\) is linear, not quadratic. Exam tip: always expand and collect like terms to write the equation as \(ax^2+bx+c=0\) before choosing the answer.

Multiply each term: \((x+4)(x+5)=x\cdot x + x\cdot5 + 4\cdot x + 4\cdot5 = x^2+5x+4x+20\). Combining like terms gives \(x^2+9x+20=0\), so option A is correct. Distractors fail due to sign or coefficient errors (D has the constant as \(-20\); B and C have incorrect coefficients/signs). Exam tip: use FOIL (first, outer, inner, last) or multiply term-by-term and then combine like terms to avoid sign mistakes.

Multiply the factors: \((x-6)(x+2)=x^2+2x-6x-12=x^2-4x-12\). Thus the standard form is \((x^2-4x-12=0)\). Option B has the wrong sign on the x-term (+4x instead of -4x); option C shows -8x which results from incorrect addition of like terms. Exam tip: Use FOIL or multiply term-by-term and then combine like terms carefully.

A coefficient is the number multiplied by a variable term. In the expression \\(x^2+12x+36=0\\), the squared term is written as \\(x^2\\). When a variable or its power appears without a visible number before it, the hidden multiplier is 1. Thus, \\(x^2\\) really means \\(1x^2\\), so its coefficient is 1. The coefficient is not 12, because 12 belongs to the \\(x\\) term, and it is not 36, because 36 is the constant term.

Therefore, option B is correct. A useful way to identify coefficients is to separate the terms: \\(x^2\\) has coefficient 1, \\(12x\\) has coefficient 12, and \\(36\\) has coefficient 36 as a constant. The equation sign and the value on the other side do not change the coefficient of the squared term. Recognizing an unwritten coefficient of 1 is an important basic algebra skill.

In the standard form \(ax^{2}+bx+c=0\), \(a\) is the coefficient of \(x^{2}\). In \( -2x^{2}+5x-1=0 \) the coefficient of \(x^{2}\) is \(-2\), so \(a=-2\). Option A (2) is a common trap that ignores the negative sign. Exam tip: write the equation in the form \(ax^{2}+bx+c=0\) first and read off the coefficients including their signs.

\(x^2+10x+25\) is a perfect square trinomial of the form \((a+b)^2=a^2+2ab+b^2\). Taking \(a=x\) and \(b=5\) gives \((x+5)^2=x^2+2\cdot x\cdot5+5^2=x^2+10x+25\), so the factorised form is \((x+5)^2=0\). The closest distractor \((x-5)^2\) is incorrect because \((x-5)^2=x^2-10x+25\), which has \(-10x\) instead of \(+10x\). Exam tip: compare the middle term with \(2ab\) and check the sign to identify the correct binomial square.

Factor the quadratic: \(x^2-3x-10=(x-5)(x+2)\). Thus the roots are \(x=5\) and \(x=-2\). Checking \(5\): \(5^2-3\cdot5-10=25-15-10=0\), so \(x=5\) is a solution. A common distractor is \(2\); checking shows \(2^2-3\cdot2-10=4-6-10=-12\), not zero, so it is not a root. Exam tip: try factoring first; if factoring is hard use substitution or the quadratic formula to verify candidates quickly.

Core concept: The degree of an equation/polynomial is the highest exponent of the variable. If that highest exponent is \(2\), the equation is called a quadratic equation and is usually written as \(ax^2+bx+c=0\). Option A is wrong because linear equations have degree \(1\). Option C is wrong because cubic equations have degree \(3\). Option D refers to a constant (no variable), which has degree \(0\). Exam tip: Always simplify and write the expression in standard polynomial form, then read off the largest exponent to determine the degree.

A quadratic term is the term where the variable has exponent 2, i.e. contains \(x^2\). Option C explicitly has \(4x^2\), so it is correct. The other options contain \(10x\), \(2x\) and \(x\), which are linear (degree 1) terms and not quadratic. The closest possible confusion could be mistaking a large coefficient (like 10 in option A) for a squared term, but the exponent must be 2. Exam tip: always check the exponent on x — only an explicit \(x^2\) (or equivalent) makes the term quadratic.

The linear term is the term containing only \(x\) (power 1). In the standard form \(ax^2+bx+c\), the linear term is \(bx\). Here \(b=0\), so the linear term is \(0x\). The closest distractor, \(9x\), would be correct only if \(b=9\), which is not the case; \(9x^2\) is the quadratic term and \(2\) is the constant term. Exam tip: write the quadratic as \(ax^2+bx+c\) and read off \(b\); if \(b=0\), the linear term is \(0x\) or is omitted in the written expression.

Since \((x-3)^2 = x^2-6x+9\), the equation \((x-3)^2=0\) is equivalent to the given quadratic. It has a repeated root \(x=3\). The closest distractor \((x+3)^2=0\) expands to \(x^2+6x+9\), so the sign of the linear term is wrong and that option is incorrect. Exam tip: Recognise perfect-square trinomials by checking if the constant equals the square of half the linear coefficient (here \((-6/2)^2=9\)).

\(x^2-49\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\) with \(a=x, b=7\) gives \((x-7)(x+7)\), so the factorised equation is \((x-7)(x+7)=0\). Option B expands to \(x^2-48x-49\), so its middle term is incorrect. Options C and D expand to \(x^2+14x+49\) and \(x^2-14x+49\) respectively, which do not match the original expression. Exam tip: recognize the difference-of-squares pattern quickly and apply \((a-b)(a+b)\).

Substituting \(x=2\) gives \(2^2+5\cdot2-14=4+10-14=0\). Hence the left-hand side equals 0, so \(x=2\) is a root of the quadratic. Option C (4) is wrong because it corresponds to evaluating only \(x^2\); option D (-4) is the value of \(5x-14\) alone. Exam tip: substitute values and simplify step by step (evaluate each term first), then combine to avoid sign or arithmetic mistakes.