Introduction to Quadratic Equations

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 product of roots is (\frac{12}{q}=4), so (q=3). Use (\frac{c}{a}) for the product of roots.

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.

For equal roots, (D=0), so (324-4k=0) and (k=81). For equal roots, set the discriminant to zero.

For equal roots, (k^2=324), and the equal root is (-\frac{k}{2}). For a negative root, (k=18) is correct.

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.