Introduction to Quadratic Equations

In (2x^2+7x=0), the (x^2) term is present and the constant term is absent. An equation can be quadratic even without a constant term.

Dividing every term by (8) gives (x^2-4x+3=0). Dividing by a common nonzero factor does not change the roots.

The numbers 8 and 9 have sum \(8+9=17\) and product \(8\times9=72\). Hence, \(x^2-17x+72=(x-8)(x-9)\). Although 6 and 12 have product 72, their sum is 18, so they do not work. Exam tip: For \(x^2+bx+c\), look for two numbers whose sum is \(b\) and whose product is \(c\).

For the quadratic equation \(x^2+vx+28=0\), the sum of the roots is \(-v\). The given roots have sum \(-4+(-7)=-11\). Hence, \(-v=-11\), so \(v=11\). Option \(-11\) is the sum of the roots, not the value of \(v\). Exam tip: For \(ax^2+bx+c=0\), the sum of roots is \(-b/a\).

If the side of the square is \(x\), its area is \(x^2\). The statement says that the area is 21 more than \(10x\), so \(x^2=10x+21\). Moving all terms to one side gives \(x^2-10x-21=0\). In option B, the sign of the \(10x\) term is incorrect. Exam tip: Translate “more than” by adding to the stated quantity.

\((x-9)^2=x^2-2\times9\times x+9^2=x^2-18x+81\). Hence, the given equation is equivalent to \((x-9)^2=0\). In contrast, \((x+9)^2\) has the middle term \(+18x\), so it is not correct. Exam tip: use the identity \(x^2-2ax+a^2=(x-a)^2\).

\((3x+4)^2=9x^2+24x+16\). Therefore, \(9x^2+24x+16=7x+16\). Moving all terms to the left gives \(9x^2+24x+16-7x-16=0\), so \(9x^2+17x=0\). Option B incorrectly treats \(24x-7x\) as \(31x\). Exam tip: write every quadratic equation in the form \(ax^2+bx+c=0\) and combine like terms carefully.

A root makes the equation equal to zero. Substituting \(x=4\) gives \(4^2+4s-32=0\), or \(16+4s-32=0\). Hence, \(4s=16\), so \(s=4\). The value \(-4\) does not satisfy the equation for \(x=4\). Exam tip: Substitute the given root directly into the equation to find an unknown coefficient.

Here (a=7), (b=-5), (c=2), so (b^2+ac=25+14=39). In (b^2), the negative sign becomes positive after squaring.

In (x^2-36=0), (b=0), so the sum of roots is (-\frac{b}{a}=0). If the (x) term is absent, the sum can be (0).

The area of a right triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Therefore, \(\frac{1}{2}x(x+2)=24\). Multiplying both sides by 2 gives \(x(x+2)=48\). On expanding, we get \(x^2+2x-48=0\), so option A is correct. Option B incorrectly misses the step of multiplying by 2. Exam tip: In area-based questions, carefully account for the \(\frac{1}{2}\) in the triangle-area formula.

A quadratic equation in x must have a non-zero coefficient of x². In the given equation, that coefficient is n² − 16. Therefore, the necessary condition is n² − 16 ≠ 0. Solving the equality that must be avoided gives n² = 16, so n = 4 or n = −4. Hence both values must be excluded, which can be written compactly as n ≠ ±4. Option A excludes only n = 4 and still permits −4, while option B excludes only −4 and still permits 4; both are incomplete. If option D were used, the x² coefficient would become zero and the equation would reduce to −3x + 7 = 0, which is linear. Thus option C is the only complete and correct answer.