Introduction to Quadratic Equations

A quadratic equation can have at most (2) real solutions. In easy questions, degree indicates the maximum possible solutions.

Match the equation to the standard form \(ax^2+bx+c=0\): here \(a=4\), \(b=4\), \(c=1\). So \(a+b+c=4+4+1=9\). Option B (8) arises from the common mistake of omitting the constant term \(c\) and adding only \(4+4=8\); therefore 8 is incorrect. Exam tip: always write the quadratic in the form \(ax^2+bx+c=0\) first, then read off and add the coefficients carefully.

From the standard form \(ax^2+bx+c=0\) we have \(a=2\), \(b=-3\), \(c=-5\). Substituting gives \(a-b+c=2-(-3)+(-5)=2+3-5=0\). Common mistake: treating \(b\) as +3 yields \(2-3-5=-6\), which is incorrect. Exam tip: always write the signs of coefficients explicitly before substituting to avoid sign errors.

The left-hand side is the polynomial \(5x^2+2x-1\). The highest power of x present is 2 (from the term \(x^2\)), so the degree of the equation is 2 — it is a quadratic equation. Option B (degree 1) would indicate a linear equation and is therefore incorrect; option A (0) refers to a constant polynomial. Exam tip: simplify the expression first if needed and then take the largest exponent of the variable (ignore terms with zero coefficients).

Expand the factors in A: \((x-5)(x+3)=x^2+3x-5x-15=x^2-2x-15\), so A matches the given quadratic. A common distractor B expands to \((x+5)(x-3)=x^2+2x-15\) which has +2x (not -2x), so it is incorrect. Exam tip: find two numbers whose product is -15 and sum is -2 (here -5 and 3) to factor quickly.

Area = length × width. With length \(x+3\) and width \(x\), the equation is \(x(x+3)=10\), which rearranges to the quadratic \(x^2+3x-10=0\) for solving. Option C (\(2x+3=10\)) might come from a mistaken linear relation (not the area formula); options B and D do not represent the product of length and width. Exam tip: Always form the product first, then bring all terms to one side to convert to standard quadratic form before solving.

If the smaller integer is \(x\), the next consecutive integer is \(x+1\). Since the relationship given is product, the correct equation is \(x(x+1)=30\). Option B uses a sum \(x+x+1=30\) instead of a product, so it is incorrect. Option C represents \(x\) being the larger integer (product \(x(x-1)\)), which contradicts the statement that \(x\) is the smaller one. Option D, \(2x=30\), models two equal integers (\(x+x\)), not consecutive integers. Exam tip: carefully note whether the problem states "product" or "sum" and which integer (smaller/larger) is denoted by the variable.

The correct standard form is \(x^2 - 5x + 6 = 0\). To convert, bring all terms to one side and set the equation equal to zero; moving \(-6\) from the right to the left changes its sign to \(+6\). Option (B) shows the common mistake where the sign of \(-6\) was not changed, so it is the wrong conversion and therefore the required 'common mistake'. The closest distractor (D) simply leaves the equation unchanged and is not in standard form; (C) unnecessarily writes \(0x\), which is not the standard conversion. Exam tip: always move all terms to one side, flip signs when moving terms, combine like terms and set equal to zero.

Putting (x=0) in option (A) gives (0=0). Direct substitution is the simplest way to check a solution.

The governing concept is comparison with the standard quadratic form ax² + bx + c = 0. In this form, a is the coefficient of x², b is the coefficient of x, and c is the constant term. In the given equation, the middle term is kx. Since the coefficient is the factor multiplying the variable, k is the coefficient of x in kx. The coefficient of x² is 1, while 12 is the constant term because it contains no variable. The zero on the right is the value to which the expression is equated, not the term represented by k. Therefore, k is the coefficient of the x term, so option B is correct. The other choices confuse the roles of a, b, and c.

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.