Introduction to Quadratic Equations

The discriminant is given by \(D=b^2-4ac\). Here \(a=4,\; b=1,\; c=6\). So \(D=1^2-4\cdot4\cdot6=1-96=-95\). Since \(D<0\), the discriminant is negative; the equation has no real roots (roots are complex). The closest distractor 'zero' is wrong because that would require \(b^2=4ac\), which is not true here. Exam tip: compute \(4ac\) carefully and watch signs to avoid arithmetic errors.

Monic means the coefficient of \(x^2\) is 1. For roots \(-5\) and \(4\) the factors are \((x+5)\) and \((x-4)\). Multiplying gives \((x+5)(x-4)=x^2+x-20\), so the monic quadratic is \(x^2+x-20=0\). The closest distractor is B, which has the wrong sign on the middle term; since the sum of the roots is \(-1\), the middle coefficient must be \(-\text{(sum) }=1\), not \(-1\). Exam tip: For roots \(\alpha,\beta\) use \(x^2-(\alpha+\beta)x+\alpha\beta=0\) to write the monic equation quickly.

The sentence means "subtract 9x from x^2 and the result is 36", so it gives \(x^2-9x=36\). Moving all terms to one side yields \(x^2-9x-36=0\), which is option A. Option B is wrong because it has the wrong sign on the linear term (it treats subtraction as addition). Option C wrongly changes the coefficient of \(x^2\), and option D swaps the numeric roles of 9 and 36; neither matches the given statement. Exam tip: translate the sentence into an equation step by step, then bring all terms to one side to form a standard quadratic equal to zero.

Take the smaller integer as \(x\); the two consecutive integers are \(x\) and \(x+1\). Their squares sum to \(x^2+(x+1)^2=145\). Expanding gives \(x^2+x^2+2x+1=145\) or \(2x^2+2x+1=145\). Subtracting 145 yields \(2x^2+2x-144=0\), which is option A. The closest distractor, B, is wrong because it ignores the \(+1\) from \((x+1)^2\) and thus miscalculates the constant term. Exam tip: divide the equation by 2 to get \(x^2+x-72=0\), factor to \((x+9)(x-8)=0\) and choose the positive root \(x=8\).

The breadth is \(x\) units, so the length is \(x+7\) units. Using area, \(x(x+7)=120\). On expanding, \(x^2+7x=120\), so \(x^2+7x-120=0\). Option B incorrectly subtracts 7, whereas the length is 7 more than the breadth. Exam tip: In rectangle word problems, first express both dimensions and set their product equal to the area.

Factor the polynomial: \(x^2-10x = x(x-10)\). By the zero-product property, \(x(x-10)=0\) implies \(x=0\) or \(x-10=0\), giving roots \(0\) and \(10\). Thus (0, 10) is correct. Option B wrongly takes the second root as −10 (sign error); options C and D reflect incorrect factorisation or arithmetic. Exam tip: factor first; then set each factor equal to zero to find the roots quickly and avoid sign mistakes.

(49x^2=64), so (x^2=\frac{64}{49}) and (x=\pm\frac{8}{7}). Take both signs while taking square roots.

For the quadratic, the constant term is 30 and the coefficient of x is 11. Find two numbers whose product is 30 and whose sum is 11 — these are 5 and 6 because \(5\cdot6=30\) and \(5+6=11\). Hence the correct factorization is \((x+5)(x+6)=0\). The closest distractor \((x+3)(x+10)=0\) has the correct product but a sum of \(3+10=13\), so it does not match the middle coefficient. Exam tip: to factor quickly, match the product (constant) and the sum (middle coefficient) before expanding.

For a quadratic equation \(ax^2+bx+c=0\), \(ac\) is the product of coefficients \(a\) and \(c\). Here \(a=4\) and \(c=3\), so \(ac=4\times3=12\). Option B (13) is incorrect because 13 is the value of \(b\), not \(ac\); option C (16) would be wrong if someone mistakenly took both coefficients as 4. Exam tip: always identify \(a, b, c\) explicitly before computing any derived quantity.

Multiplying the whole equation by (5) gives (x^2+10x-35=0). To remove fractions, multiply the whole equation.

Denominators 4 and 2 have LCM = 4, so multiply the entire equation by 4: \(\frac{3}{4}x^2\times4=3x^2,\; -\frac{1}{2}x\times4=-2x,\; 2\times4=8\). Thus the integer-coefficient form is \(3x^2-2x+8=0\). Option B is incorrect because the constant term was not multiplied correctly (it should become 8). Exam tip: to remove fractional coefficients, multiply the whole equation by the LCM of all denominators — do not change individual terms differently.

(\frac{1}{x^2}=x^{-2}), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.

In the standard form \(ax^2+bx+c=0\), \(b\) is the coefficient of \(x\). In the given equation the coefficient of \(x\) is \(6\sqrt{5}\), so \(b=6\sqrt{5}\). Option B is incorrect because it has the wrong sign; option C is the constant term \(c\); option D equals the coefficient \(a\). Exam tip: always write the quadratic in standard form and read off the coefficients for \(a, b, c\).

The product of the roots is simply their multiplication: \\((-4)(-9)=36\\). The product of two negative numbers is positive, so 36 is correct. The closest distractor \\(-36\\) is wrong due to an incorrect sign; \\(13\\) and \\(-13\\) confuse product with the sum (the sum is \\(-13\\)). Exam tip: check signs first — if both roots are negative, the product is positive.

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-8) gives (x^2+8x+15=0).

The sum of roots is (-\frac{b}{a}=-\frac{-9}{5}=\frac{9}{5}). Pay attention to the sign of (b) in the formula.

The sum of roots of a quadratic equation ax^2 + bx + c = 0 is -b/a. In 4x^2 + mx + 20 = 0, a = 4 and b = m, so the sum is -m/4. The question gives this sum as -6. Equating these two expressions provides a direct linear equation for m. The constant term 20 is irrelevant because it is used for the product, not the sum.

Set -m/4 = -6. Multiplying by 4 gives -m = -24, and changing the signs gives m = 24. Therefore option A is correct. The answer would be -24 only if the leading negative sign in the root-sum formula were mishandled. Options C and D confuse the value of the root sum or its coefficient with the requested coefficient m.

The direct answer is A: \(q=3\). For \(ax^2+bx+c=0\), the product of roots is \(c/a\). Here \(a=q\) and \(c=12\), so the product is \(12/q\). The question says this product is 4, hence \(12/q=4\). Multiplying by \(q\) gives \(12=4q\), and dividing by 4 gives \(q=3\). Option A is correct. Option B, 4, would give product \(12/4=3\), not 4. Option C, 6, gives product 2. Option D, -3, gives product -4, including the wrong sign. The equation remains quadratic for \(q=3\), since its leading coefficient is nonzero. Memory cue: product of roots is constant term divided by the coefficient of \(x^2\).

Factor the quadratic: \(x^2-3x-40=(x-8)(x+5)\). Thus the roots are \(x=8\) and \(x=-5\). Verify by sum and product: sum \(8+(-5)=3\) equals \(-b/a=-(-3)/1=3\), and product \(8\times(-5)=-40\) equals \(c/a=-40\). The closest distractor \((5,-8)\) has the correct product but wrong sum (\(-3\) instead of \(+3\)), so it is incorrect. Exam tip: use sum = \(-b/a\) and product = \(c/a\) to check root choices quickly.

If both roots are 8, their sum is \(8+8=16\). For \(ax^2+bx+c=0\) the sum of roots equals \(-b/a\). Here \(a=1, b=u\), so \(-u=16\) implying \(u=-16\). Alternatively, discriminant zero gives \(u^2-4\cdot1\cdot64=0\Rightarrow u=\pm16\), and matching the given root value yields \(u=-16\). Exam tip: substitute the root or expand \((x-8)^2\) to check your value quickly.

Expand terms: \((x+4)^2 = x^2+8x+16\) and \(-3(x+4) = -3x-12\). Adding these and \(-10\) gives \(x^2+8x+16-3x-12-10 = x^2+(8x-3x)+(16-12-10) = x^2+5x-6\). Thus the quadratic is \(x^2+5x-6=0\). Closest distractor (option C) has the constant term sign reversed — a typical error from mishandling the constant terms. Exam tip: always expand brackets first and then combine like terms separately to avoid sign mistakes.

Expanding gives \((x-4)(x+9)=x^2+9x-4x-36=x^2+5x-36\). Now subtract \(5x\) from both sides: \(x^2+5x-36-5x=0\), so \(x^2-36=0\). Option B is only the expanded left-hand side and does not account for the \(5x\) on the right. Exam tip: To write a quadratic in standard form, bring all terms to one side and make the other side \(0\).

Substituting \(x=7\), the left-hand side becomes \(7^2-15\times7+56=49-105+56=0\). Hence, the correct answer is 0, and \(7\) is a root of the quadratic equation. Note that 7 is the value substituted for \(x\), not the value of the left-hand side. Exam tip: To verify a root, substitute it into the left-hand side; the result must be 0.

Direct answer: Option B, k=81. Equal roots require discriminant \\(D=0\\). In \\(x^2-18x+k=0\\), a=1, b=-18, c=k. Therefore \\(D=(-18)^2-4(1)(k)=324-4k\\). Put this equal to zero: \\(324-4k=0\\); then \\(4k=324\\), so \\(k=81\\). Option B is correct. Option A, 18, does not satisfy the equation because its discriminant is not zero. Option C, 162, and option D, 324, also leave nonzero discriminants and therefore do not produce equal roots. A useful check is \\(x^2-18x+81=(x-9)^2\\), showing the repeated root is 9. Remember: equal roots means b²−4ac must be exactly zero.

The direct answer is option A, \(k=18\). For \(x^2+kx+81=0\), equal roots require the discriminant to be zero: \(k^2-4(1)(81)=0\), so \(k^2=324\) and \(k=18\) or \(k=-18\). The repeated root is \(-k/2\). The question says the equal roots are negative. If \(k=18\), the root is \(-18/2=-9\), which is negative. If \(k=-18\), the root is \(9\, ext{, positive}\), so it does not satisfy the condition. Thus option A is correct. Option B gives equal roots but they are positive, not negative. Option C, 9, gives discriminant \(81-324\ne0\), so roots are not equal. Option D, −9, also gives a non-zero discriminant and unequal roots. Memory cue: equal roots mean discriminant zero; then check the sign of the repeated root.