Roots of a Quadratic Equation

Use the formula \(D=b^2-4ac\). Here \(a=1,\; b=-2,\; c=5\), so \(D=(-2)^2-4\cdot1\cdot5=4-20=-16\). Thus the correct discriminant is -16; a negative D means no real roots (the roots are complex conjugates). The closest distractor 16 is wrong because it comes from treating b as +2 instead of -2. Exam tip: always substitute b with its sign (the coefficient as written) to avoid sign errors.

The sign of the discriminant D determines the nature of the roots. When \(D<0\) the quantity \(\sqrt{D}\) is not real, so the roots are complex conjugates and there are no real roots. If \(D>0\) there are two distinct real roots, and if \(D=0\) there is one repeated real root; thus those options are incorrect. Exam tip: compute \(b^2-4ac\) quickly — if it is negative you can immediately conclude there are no real roots.

For any real \\((x)\\), \\(x^2\\ge0\\), so \\(x^2+9\\ge9>0\\) and cannot be zero. The discriminant \\(b^2-4ac = 0^2 - 4\cdot1\cdot9 = -36 < 0\\) therefore confirms there are no real roots; the equation has complex roots \\(\pm 3i\\). Closest distractor (option D) is wrong because \\(3^2+9=18\\), not zero. Exam tip: check the discriminant \\(b^2-4ac\\) to decide existence of real roots quickly.

Factor the quadratic: \(x^2-5x = x(x-5)\). If a product is zero, at least one factor is zero, so \(x=0\) or \(x-5=0\) giving \(x=5\). Thus the roots are 0 and 5. The closest distractor is D (0 and -5) — it has 0 correct but the sign of the other root is wrong because the factor is \(x-5\), not \(x+5\). Exam tip: Always factor and set each factor to zero; carefully check signs when solving for roots.

Write the equation as \(x^2-9x=0\) and factor: \(x(x-9)=0\). By the zero-product property either \(x=0\) or \(x-9=0\Rightarrow x=9\). Hence the roots are 0 and 9. A common mistake is dividing both sides by \(x\), which loses the root \(x=0\). Exam tip: bring to standard form and factor, or if you divide by a variable, always check for the lost zero root.

Divide both sides by 3: \(x^2=9\). Taking square roots gives \(x=\pm3\), so the roots are 3 and −3. Option B (9, −9) is incorrect because it confuses the value of \(x^2\) with the values of x (the square roots are ±3, not ±9). Options C and D omit the negative root. Exam tip: first divide by the leading coefficient to get \(x^2=\) form, then apply ±√ to find both roots.

Factor the quadratic: \(x^2-8x+16=(x-4)^2\). Hence \((x-4)^2=0\) gives the repeated root \(x=4\). Alternatively compute the discriminant: \(\Delta=b^2-4ac=(-8)^2-4\cdot1\cdot16=0\), and \(\Delta=0\) indicates a double root. The closest distractor \(-4\) is incorrect because substituting \(-4\) does not satisfy the equation (it gives 64). Exam tip: either factor into a perfect square or check the discriminant; if \(\Delta=0\) the root is repeated.

If a quadratic has roots α and β, it can be written as \((x-α)(x-β)=0\). With roots -2 and 6 we get \((x+2)(x-6)=0\). Expanding gives \(x^2-4x-12=0\), so option A is correct. For comparison, option C expands to \((x-2)(x-6)=x^2-8x+12\) which has roots 2 and 6, so it is wrong. Exam tip: check sum and product of roots quickly — for a monic quadratic \(x^2+bx+c\), sum of roots = -b and product = c; here sum = -2+6=4 (so \(b=-4\)), product = -12.

Use factorisation or Vieta's relations. Factorising gives \(x^2-9x+18=(x-3)(x-6)\), so if one root is 3 the other is 6. By Vieta, the sum of roots is \(\alpha+\beta=9\); with \(\alpha=3\) we get \(\beta=6\). The distractor 3 is incorrect because a repeated root 3 would make the polynomial \((x-3)^2=x^2-6x+9\), which does not match the given polynomial. Exam tip: try quick factorisation first; if not obvious use sum/product of roots.

A positive product means both roots have the same sign. A negative sum means both roots are negative.

If \(x+5\) is a factor, set the factor equal to zero: \(x+5=0\) gives \(x=-5\). Hence the certain root is \(-5\). Option A (5) is incorrect because the sign is reversed; solving \(x+5=0\) does not give +5. Option C (0) would only be correct if the factor were \(x\). Option D (\(\tfrac{1}{5}\)) would arise from a factor \(x-\tfrac{1}{5}\), not \(x+5\). Exam tip: always equate the linear factor to zero to find its root quickly.

Substitute \(x=4\) into the equation: \(16+4m-20=0\), so \(4m-4=0\) and hence \(m=1\). The nearest distractor (\(-1\)) is wrong because with \(m=-1\) the left side becomes \(16-4-20=-8\), not zero. Exam tip: you can also use Vieta — with one root 4 the other is \(-5\) (product = \(-20\)), sum = \(4+(-5)=-1=-m\) giving \(m=1\).

Dividing \(7x^2=0\) by 7 gives \(x^2=0\). This is \((x-0)^2\), so \(x=0\) is a repeated root (multiplicity 2). Using coefficients \(a=7,b=0,c=0\), the discriminant is \(b^2-4ac=0^2-4\cdot7\cdot0=0\), confirming a double root. The closest distractor C is wrong because substituting \(x=7\) gives \(7\cdot7^2=343\ne0\). Exam tip: check the discriminant or factor out common terms to quickly identify repeated roots.

This is a difference of squares since \(x^2-49=(x-7)(x+7)\). Setting each factor to zero gives roots: from \(x-7=0\) we get \(x=7\), and from \(x+7=0\) we get \(x=-7\). Option B is wrong because using 49 or −49 would imply \(x^2=2401\), which does not satisfy the original equation. Exam tip: when taking square roots remember both signs — from \(x^2=a\) we get \(x=\pm\sqrt{a}\).

Factor the expression: \(x^2+4x = x(x+4)\). By the zero-product property, the roots are \(x=0\) and \(x=-4\). Among the given choices \(-4\) is present, so it is correct. For example, substituting \(x=4\) gives \(4^2+4\cdot4=16+16=32\neq0\), so 4 is not a root. Exam tip: always try factoring out the common factor first to apply the zero-product rule quickly.

For a monic quadratic with roots r1 and r2 use \(x^2-(r1+r2)x+r1r2=0\). Here r1=5 and r2=−3 so sum = \(5+(-3)=2\) and product = \(5\times(-3)=-15\). Thus the equation is \(x^2-2x-15=0\) (option A). The closest distractor is option C, \(x^2-8x+15=0\), whose sum is 8 and product 15, giving roots 3 and 5 — it shares root 5 but the other root is different. Exam tip: compute sum and product of given roots and form \(x^2-( ext{sum})x+( ext{product})=0\).

For a quadratic \(ax^2+bx+c=0\), the sum of the roots equals \(-\frac{b}{a}\). Here \(a=1\) and \(b=11\), so the sum is \(-\frac{11}{1}=-11\). Option A (11) is incorrect because the sign must be negated; options C and D correspond to the constant term 30 (or its negative) and are not the sum. Exam tip: quickly identify a and b and apply \(-\frac{b}{a}\).

For a quadratic \(ax^2+bx+c=0\) the product of roots is \(\alpha\beta=\dfrac{c}{a}\). Here \(a=5\) and \(c=-8\), so the product is \(-\dfrac{8}{5}\). Option B has the wrong sign; options C and D are incorrect numerical values. Exam tip: identify \(a\) and \(c\) first and directly compute \(c/a\) to avoid sign mistakes.

If \(x=0\) is a root, substitute \(x=0\) into \(2x^2+bx+c\). That gives \(2(0)^2+b(0)+c=c\), so \(c=0\). Option C (\(c=b\)) is incorrect because the coefficient \(b\) need not equal the constant term without extra conditions. Options B and D are arbitrary numeric values and are not determined by the given information. Exam tip: For a zero root, check the constant term first—it must be zero.

Putting (x=-1) gives (1-r-12=0), so (r=-11). Check the sign of each term while using a negative root.