Introduction to Quadratic Equations

For a quadratic to have real and distinct roots the discriminant \(D=b^2-4ac\) must be positive. Here \(a=1,\;b=-20,\;c=k\), so
\(D=(-20)^2-4\cdot1\cdot k=400-4k\).
Real and distinct roots require \(D>0\), hence \(400-4k>0\) which gives \(k<100\).
If \(k=100\) the roots are equal, and if \(k>100\) the roots are complex. Exam tip: factor common multiples (\(D=4(100-k)\)) to check the sign quickly.

Discriminant is defined by \(D=b^2-4ac\). For option A, \(a=1,\;b=1,\;c=12\), so \(D=1^2-4\cdot1\cdot12=1-48=-47\), which matches. The closest distractor, option D, gives \(D=(-8)^2-4\cdot1\cdot30=64-120=-56\), not -47. Exam tip: always identify \(a,b,c\) first and compute \(b^2-4ac\) carefully, watching signs of b and c.

\(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\). Here \(\left(-\frac{7}{6}\right)^2-2\cdot\frac{1}{3}=\frac{25}{36}\).

Use the identity \((\alpha-8)(\beta-8)=\alpha\beta-8(\alpha+\beta)+64\). For \(x^2-18x+80=0\), by Vieta \(\alpha+\beta=18\) and \(\alpha\beta=80\). Therefore the value is \(80-8\times18+64=80-144+64=0\). A common mistake is to pick \(80\) (the product \(\alpha\beta\)) and ignore the linear shift terms. Exam tip: Apply Vieta's formulas directly or set \(y=x-8\) to transform the expression quickly.

The standard form of a quadratic equation is written as a polynomial equal to zero. First expand each product carefully. We get \\(x+5)(x+8)=x^2+13x+40\\) and \\(x-4)(x+7)=x^2+3x-28\\). Adding them gives \\(2x^2+16x+12\\), so the left side is a quadratic expression.

The original equation says this expression equals 110. Move 110 to the left by subtracting it from both sides: \\(2x^2+16x+12-110=0\\). Combining the constants gives \\(2x^2+16x-98=0\\). Therefore, option A is correct. Option C loses a factor of 2, while option D does not correctly expand the products or rearrange the constant.

From (3r+4r=28), we get (r=4), so the roots are (12) and (16). The constant term is the product of roots (192).

Compute the discriminant: \(D=b^2-4ac = (-12)^2 - 4\cdot1\cdot40 = 144 - 160 = -16 < 0\). When \(D<0\) a quadratic has no real roots (the roots are non-real complex conjugates). Thus option A is correct. Option B is wrong because equal (repeated) real roots occur only if \(D=0\). Option C is wrong because two distinct real roots require \(D>0\). Option D is wrong because the product of the roots equals \(c/a = 40\), not \(-40\). Exam tip: always evaluate \(D\) first to decide the nature of roots; use sum = \(-b/a\) and product = \(c/a\) for quick checks.

For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(c/a\). Here \(a=1\) and \(c=7m\), so the product is \(7m\). Given one root is 7, the other root is \(\dfrac{7m}{7}=m\). Check with the sum: sum = \(-b/a\) = \(m+7\), which matches \(7+m\). The closest distractor (option B: \(7m\)) confuses the product with a root. Exam tip: always apply sum = \(-b/a\) and product = \(c/a\), and watch the signs of coefficients.

Compute the discriminant: for \(7x^{2}-30x+32=0\), \(a=7,\;b=-30,\;c=32\) so \(D=b^{2}-4ac=900-896=4\). The difference between roots equals \(|r_{1}-r_{2}|=\dfrac{\sqrt{D}}{a}\). Therefore the difference is \(\dfrac{\sqrt{4}}{7}=\dfrac{2}{7}\). Note: the option \(\tfrac{4}{7}\) results from mistakenly using \(D/a\), and \(\tfrac{16}{7}\) is one of the roots (the other root is 2), not their difference. Exam tip: use \(|r_{1}-r_{2}|=\sqrt{D}/|a|\) to get the answer quickly without finding both roots explicitly.

A perfect-square trinomial must match \((x+b)^2 = x^2 + 2bx + b^2\). Here the coefficient of \(x\) is 24, so \(2b=24\) giving \(b=12\), and hence \(c=b^2=12^2=144\). Alternatively use discriminant: \(b^2-4ac=24^2-4\cdot1\cdot c=576-4c=0\) which yields \(c=144\). Note: 576 might look tempting because it is \(24^2\), but the constant term must be \(b^2\) with \(2b=24\), not \(24^2\). Exam tip: either complete the square or set the discriminant to zero to solve such questions quickly.

For a quadratic, the sum of roots is \(\alpha+\beta=-b/a\). Here \(a=1\) and both roots are \(-12\), so the sum is \(-24\). Thus \(-b=-24\) which gives \(b=24\). Alternatively, for a repeated root use \(-b/(2a)\): \(-b/2=-12\) leads to the same result. The distractor \(-24\) is the sign-error mirror of the correct sum, so it is incorrect. Exam tip: use \(\alpha+\beta=-b/a\) and \(\alpha\beta=c/a\) and double-check signs when substituting.

In the first option, the sum is (-\frac{b}{a}=9) and the product is (\frac{c}{a}=20). So both are positive.

Area gives \((x+10)(x-7)=160\). Expanding the left side yields \(x^2+3x-70=160\). Bringing 160 to the left gives \(x^2+3x-70-160=0\), i.e. \(x^2+3x-230=0\), so option A is correct. The closest wrong option is B, which has the constant term \(-160\) instead of \(-230\) — a typical mistake from not moving the right-hand term across correctly. Exam tip: first expand, then move all terms to one side to form the quadratic and carefully check signs and arithmetic.

If the sum of roots is (19) and product is (90), the equation is (x^2-19x+90=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{17}{72}).

Using the identity \((x+a)^2 = x^2 + 2ax + a^2\), compare coefficients with \(x^2 + 28x + 196\). From the coefficient of \(x\) we get \(2a = 28\), so \(a = 14\). Also \(a^2 = 196\) holds for \(a=14\). The option \(-14\) is incorrect because it would give \(2a = -28\), which contradicts the given linear coefficient. Exam tip: equate coefficients of like powers of \(x\) to find unknown parameters quickly.

The quadratic is a perfect square: \((6x-5k)^2=0\). Alternatively compute the discriminant: \(\Delta=b^2-4ac=(-60k)^2-4\cdot36\cdot25k^2=3600k^2-3600k^2=0\). A zero discriminant means the two roots are equal and real (root \(x=5k/6\)). Option B is wrong because that requires \(\Delta>0\); option C is wrong because that requires \(\Delta<0\); option D is wrong because the nature does not depend on a special value of k for real k — it is always equal roots. Exam tip: check the discriminant first or try factoring into a perfect square to save time.

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (64), so it is not possible.

For equal roots, (p^2-196=0) gives (p=\pm14). For equal negative roots, (-\frac{p}{2}<0), so (p=14).

Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, (9k-19\neq0), so (k\neq\frac{19}{9}).