Introduction to Quadratic Equations

Substitute the given root into the equation: \(2(3)^2+q(3)-15=0\) ⇒ \(18+3q-15=0\) ⇒ \(3+3q=0\) ⇒ \(q=-1\). Option A (1) often arises from a sign mistake when moving terms; options C and D come from arithmetic or sign errors. Exam tip: always substitute the root first and simplify step by step, checking basic addition/subtraction carefully.

(4x^2+1=3x) can be written as (4x^2-3x+1=0). In standard form, all terms are on one side and (0) is on the other.

((x+4)^2=x^2+8x+16), and subtracting (5x+10) gives (x^2+3x+6=0). Keep the middle term correct in square identities.

Substituting into (ax^2+bx+c=0) gives (5x^2-11x-6=0). Treat the negative sign as part of the coefficient.

The discriminant is given by \(D=b^2-4ac\). Here \(a=9,\; b=-12,\; c=4\). Thus \(D=(-12)^2-4\cdot9\cdot4=144-144=0\). Option B mistakenly gives only \(b^2\) (144) without subtracting \(4ac\). Option C arises from incorrectly treating \((-12)^2\) as -144; squaring a negative yields a positive. Exam tip: identify \(a,b,c\) first, then compute \(b^2-4ac\) carefully.

The direct answer is B: two equal real roots. A quadratic has the form \(ax^2+bx+c=0\). Here \(a=1\), \(b=-10\), and \(c=25\). Its discriminant is \(D=b^2-4ac=(-10)^2-4(1)(25)=100-100=0\). When \(D=0\), the two roots are real and equal; in fact, the equation is \((x-5)^2=0\), so both roots are 5. Option A is wrong because distinct real roots require \(D>0\). Option B is correct because \(D=0\). Option C is wrong because no real roots occur when \(D<0\). Option D is wrong because a quadratic has at most two roots, not three. Memory cue: positive, zero, negative discriminant means distinct, equal, and non-real roots.

The discriminant is defined by \(D=b^2-4ac\). Here \(a=3,\; b=2,\; c=7\). So \(D=2^2-4\cdot3\cdot7=4-84=-80\). The value is negative, so the discriminant is negative and the quadratic has no real roots (two complex conjugate roots). Option B (zero) is wrong because \(D=0\) would give equal real roots; option A (positive) is wrong because a positive discriminant yields two distinct real roots. Exam tip: compute \(b^2\) and \(4ac\) separately to avoid sign or multiplication errors.

For a monic quadratic with roots r1 and r2 the polynomial is \(x^2 - (r1+r2)x + r1r2 = 0\). With roots \(-4\) and \(3\), the sum is \(-1\) and the product is \(-12\). Substituting gives \(x^2 -(-1)x + (-12) = x^2 + x - 12 = 0\), so option A is correct. The closest distractor \(x^2 - x - 12 = 0\) only differs by the sign of the middle term (it assumes sum = 1 instead of -1), hence it is wrong. Exam tip: either form \((x - r_1)(x - r_2)\) and expand, or use Vieta’s relations \(\text{sum} = -b/a\) and \(\text{product} = c/a\) to check coefficients quickly.

From the statement we get \(x^2 = 5x + 24\). Bringing all terms to one side gives \(x^2-5x-24=0\), so option A is correct. Option B has the wrong sign on the middle term. Option C incorrectly places 24 as the coefficient of x. Option D wrongly multiplies \(x^2\) by 5. Exam tip: Translate "is ... more than" as equality with addition on the right, then move all terms to one side to form the quadratic.

If the smaller integer is \(x\), the next is \(x+1\). So set up \(x^2+(x+1)^2=85\). Expanding gives \(x^2+x^2+2x+1=85\), i.e. \(2x^2+2x+1-85=0\), which simplifies to \(2x^2+2x-84=0\). Option B is close but has the constant wrong (\(-85\) instead of \(-84\)). Option C misses the doubled \(x^2\) term; option D has the wrong sign on the linear term. Exam tip: expand carefully, bring all terms to one side and combine like terms before comparing options.

Length = breadth + 6, so length = \(x+6\). Area = breadth × length = \(x(x+6)=72\). Rearranging gives \(x^2+6x-72=0\), so option A is correct. The closest distractor B has the sign of the linear term wrong (\(-6x\) instead of \(+6x\)); C and D have incorrect coefficients/term placement. Exam tip: assign variables to dimensions, form the area equation, then bring all terms to one side to obtain a standard quadratic for solving.

Factor the polynomial: \(x^2+9x=x(x+9)\). By the zero-product property, if \(ab=0\) then \(a=0\) or \(b=0\). So \(x=0\) or \(x+9=0\Rightarrow x=-9\). Therefore the roots are 0 and -9. Option B is the closest distractor but wrong due to the sign of 9; options C and D use incorrect numbers. Exam tip: factorise first and apply the zero-product rule to find roots quickly and reliably.

(16x^2=81), so (x^2=\frac{81}{16}) and (x=\pm\frac{9}{4}). Take both signs while taking square roots.

Method: find two numbers whose product is 20 and sum is 9. The pair 4 and 5 satisfy \(4\cdot5=20\) and \(4+5=9\), so the factorization is \((x+4)(x+5)=0\). Option B would use -4 and -5 giving sum \(-9\), so it is incorrect. Options C and D sum to 12 and 21 respectively, so they do not match the middle coefficient. Exam tip: list factor pairs of the constant term (20 = 1·20, 2·10, 4·5) and pick the pair whose sum equals the coefficient of \(x\).

For the standard form \(ax^2+bx+c=0\), we have \(a=3\) and \(c=8\) here, so \(ac=3\times8=24\). Option B (30) is incorrect and likely arises from confusing \(ac\) with \(a\times b\) (\(3 imes10\)); option C (18) would be wrong if one mistakenly read \(c\) as 6. Exam tip: always write the equation in standard form and read off \(a, b, c\) before performing calculations.

In (5x^2+20=0), the (x) term is absent, so (b=0) and (c=20>0). Treat a missing term coefficient as (0).

Multiplying the whole equation by (4) gives (x^2-4x+12=0). To remove fractions, multiply the whole equation.

To get integer coefficients, eliminate denominators by multiplying every term by the common denominator. Here the common denominator is 5. Multiplying the whole equation by 5 gives: \(5\cdot\frac{2}{5}x^2=2x^2,\;5\cdot\frac{1}{5}x=1x,\;5\cdot(-2)=-10\). So the equivalent equation is \(2x^2+x-10=0\). Closest distractors fail because they omit multiplying the constant (\(2x^2+x-2=0\)) or misapply multiplication to only some terms (\(5x^2+x-2=0\), \(2x^2+5x-10=0\)). Exam tip: always multiply every term (including the constant and both sides) by the common denominator, then simplify by any common factor if possible.

The term (\sqrt{x}) has a fractional power of the variable, so it is not in usual quadratic form. Quadratic form has only (x^2), (x), and constant terms.

In the standard quadratic form \(ax^2+bx+c=0\), b is the coefficient of x. In \(x^2-4\sqrt{2}x+8=0\) the coefficient of x is \(-4\sqrt{2}\), so b = -4\sqrt{2}. Option B (\(4\sqrt{2}\)) is the same magnitude but with the wrong sign — include the sign when identifying coefficients. Exam tip: always arrange the equation in standard form and read off a, b, c (with their signs).

Compute the product directly: (-3)×(-8)=24. The product of two negative numbers is positive, so 24 is correct. The closest distractor -24 is wrong due to sign error (making the result negative). The options 11 and -11 are incorrect because they do not match the multiplication. Exam tip: For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(c/a\); for simple numeric roots just multiply them and check the sign rule for negatives.

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-6) gives (x^2+6x+8=0).

The sum of roots is (-\frac{b}{a}=-\frac{-7}{4}=\frac{7}{4}). Pay attention to the sign of (b) in the formula.

For a quadratic equation ax^2 + bx + c = 0 with roots alpha and beta, the product of the roots is given by alpha beta = c/a. This result comes from comparing the constant term with the leading coefficient; the coefficient b is not needed for this particular question. The sign of c must be retained exactly as it appears.

Here a = 5 and c = -11 in 5x^2 - 3x - 11 = 0. Therefore, product of roots = c/a = -11/5. Hence option A is correct. A positive 11/5 would ignore the negative constant term, while -3/5 is related to b/a and is not the product of the roots. The given answer correctly applies the root-product formula.

For a quadratic equation ax^2 + bx + c = 0, the sum of its roots is -b/a. The minus sign is essential. In the equation 3x^2 + nx + 12 = 0, the leading coefficient is a = 3 and the coefficient of x is b = n. Therefore, the sum of roots is -n/3. The stated sum allows n to be found by a simple equation.

We are told that -n/3 = -4. Multiplying both sides by 3 gives -n = -12, and multiplying by -1 gives n = 12. Thus option A is correct. If the negative sign in the formula were missed, one might incorrectly obtain n = -12. The constant term 12 is not required for this calculation, because only the sum of roots is being used.