Lesson Using Vieta's theorem to solve quadratic equations and related problems
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<H2>Using Vieta's theorem to solve quadratic equations and related problems</H2> <H3>Problem 1</H3>If "a" and "b" are the roots of the quadratic equation {{{x^2 - 3x -1}}} = 0, find {{{a^2 + b^2}}}. <B>Solution 1</B> <pre> According to Viets's theorem, a + b = 3 and ab = -1. Hence, {{{a^2 + b^2}}} = {{{(a + b)^2 - 2ab}}} = {{{3^2 - 2*(-1)}}} = 9 + 2 = 11. <U>Answer</U>. {{{a^2 + b^2}}} = 11. Notice that we do not need to solve the given quadratic equation explicitly to get the answer. </pre> <H3>Problem 2</H3>If two roots of the quadratic polynomial {{{7x^2+3x+k}}} are {{{(-3 +- i*sqrt(299))/14}}}, find k. <B>Solution</B> <pre> {{{k/7}}} is the product of the roots (Vieta's theorem): {{{k/7}}} = {{{((-3+i*sqrt(299))/14)*((-3-i*sqrt(299))/14)}}} = {{{(3^2+299)/14^2}}} = {{{308/14^2}}}. Hence, k = {{{(308*7)/14^2}}} = 11. </pre> <H3>Problem 3</H3>The roots of the quadratic equation {{{z^2 + az + b}}} = 0 are 2 - 3i and 2 + 3i. What is a+b? <B>Solution</B> <pre> -a = (2-3i) + (2+3i) = 4 ====> a = -4. b = (2-3i)*(2+3i) = {{{2^2 + 3^2}}} = 4 + 9 = 13. ====> a + b = -4 + 13 = 9. The Vieta's theorem. <U>Answer</U>. a + b = 9. </pre> <H3>Problem 4</H3>Suppose "a" and "x" satisfy {{{x^2 + (a-(1/a))x - 1}}} = 0. Solve for "x" in terms of "a". <B>Solution</B> <U>Solution 1</U> (the Vieta's theorem) <pre> The Vieta's theorem says: if p and q are the roots of a quadratic equation {{{x^2 + ux + v}}} = 0 then u = -(p+q) and v = pq. The opposite is also TRUE: if u = -(p+q) and v = pq then p and q are the roots of the quadratic equation {{{x^2 + ux + v}}} = 0. Now look into your equation and notice that the numbers -a and {{{(1/a)}}} give {{{-(a - (1/a))}}} when summed up and -1 when multiplied. Hence, the numbers "-a" and {{{1/a}}} are the roots of your equation. </pre> <U>Solution 2</U> (Factoring) <pre> Factor your polynomial: {{{x^2 + (a-(1/a))x - 1 }}} = {{{(x+a)*(x-1/a)}}}. Hence, the roots are "-a" and {{{1/a}}}. </pre> <H3>Problem 5</H3>Without solving the given equation, find an equation whose roots are the squares of the roots of {{{x^2 + 4x + 2}}} = 0. <B>Solution</B> <pre> Let "s" and "t" be the roots of the quadratic polynomial p(x) = {{{x^2 + px + q}}} with the leading coefficient 1 at {{{x^2}}}. Then the Vieta's theorem says: s + t = -p and s*t = q. For the given equation {{{x^2 + 4x +2}}} = {{{0}}} it means that if "s" and "t" are its roots, then s + t = -4, (1) and s*t = 2. (2) Next, if (1) and (2) are held, then {{{s^2 + 2st + t^2}}} = {{{(s+t)^2}}} = {{{(-4)^2}}} = 16; hence, {{{s^2 + t^2}}} = 16 - 2s*t = 16 - 2*2 = 12, and {{{S^2*t^2}}} = {{{(st)^2}}} = {{{2^2}}} = 4. Hence, by applying the Vieta;s theorem once again, you see that {{{s^2}}} and {{{t^2}}} are the roots of the polynomial g(x) = {{{x^2 -12x + 4}}}. It is the answer to the problem's question. The problem is solved !! Notice, we get the answer without solving the original equation. Exactly as assigned by the condition. All we did was manipulating with coefficients and using the Vieta's theorem. </pre> <H3>Problem 6</H3>Determine all values of "c" so that (2x+1) is a factor of {{{2x^2+7x+c}}}. <B>Solution</B> <pre> 1. The fact that 2x+1 is the factor of the polynomial p(x) = 2x^2+7x+c is equivalent to divisibility the polynomial p(x) by {{{x + 1/2}}}. 2. Divisibility of the polynomial p(x) by {{{x + 1/2}}} is equivalent to the fact that {{{-1/2}}} is the root of the polynomial p(x): {{{p(-1/2)}}} = 0. (the "Remainder theorem") 3. The fact that {{{-1/2}}} is the root of the polynomial p(x) means that {{{2*(-1/2)^2 + 7*(-1/2) + c}}} = 0, or {{{2/4 - 7/2 + c}}} = 0, or c = {{{-2/4 + 7/2}}} = {{{-2/4 + 14/4}}} = {{{12/4}}} = 3. <U>Answer</U>. c = 3. </pre> <H3>Problem 7</H3>If {{{sqrt(3) - 1}}} is a root of the equation 2x^2-2kx+4 = 0, find the value of "k". <B>Solution</B> <pre> The given equation 2x^2 - 2kx + 4 = 0 is equivalent to (after dividing both sides by 2) x^2 - kx + 2 = 0. (1) We are given that {{{sqrt(3)-1}}} is the root of the original equation; hence, the equation (1) with the leading coefficient 1 has this root, too. Then, applying the Vieta's theorem, the other root of the equation (1) is {{{2/(sqrt(3)-1)}}} = {{{2/(sqrt(3)-1)}}}.{{{(sqrt(3)+1)/(sqrt(3)+1)}}} = {{{(2*sqrt(3)+1)/((sqrt(3))^2-1^2)}}} = {{{(2*(sqrt(3)+1))/2}}} = {{{sqrt(3)+1}}}. Thus we know BOTH ROOTS of the equation (1) (even without solving it explicitly (!) ). They are {{{sqrt(3)-1}}} and {{{sqrt(3)+1}}}. Again, according to Vieta's theorem, the sum of these roots is the coefficient at x in equation (1) taken with the opposite sign: k = {{{sqrt(3)-1}}} + {{{sqrt(3)+1}}} = {{{2*sqrt(3)}}}. <U>Answer</U>. k = {{{2*sqrt(3)}}}. </pre> <H3>Problem 8</H3>If the equations {{{x^2 - ax + b}}} = 0 and {{{x^2 - ex + f}}} = 0 have one root in common and if the second equation has equal roots, then prove that ae = 2(b + f). <B>Solution</B> <pre> 1. If {{{x^2 - ex + f}}} = 0 has equal roots, it means that its discriminant d = e^2 - 4f is zero: {{{e^2 - 4f}}} = 0, or {{{e^2}}} = {{{4f}}}, or {{{e^2/2}}} = 2f. (1) 2. Then the merged root of the equation {{{x^2 - ex + f}}} = 0 is {{{x[1,2]}}} = {{{e/2}}} ( <<<---=== it follows from the quadratic formula, for example ) 3. It implies that the common root of the two given equations is {{{e/2}}}, since the second equation has no other roots. 4. Thus we know that {{{e/2}}} is one of the roots of the equation {{{x^2 - ax + b}}} = 0. Then, according to the Vieta's theorem, the other root of this equation is {{{a - e/2}}}, and the product of the roots is the constant term {{{(e/2)*(a-e/2)}}} = b. 5. The last equality is equivalent to {{{ae}}} - {{{e^2/2}}} = 2b <-----> ae = {{{2b}}} + {{{e^2/2}}} <-----> ae = 2b + 2f <-----> ae = 2(b+f). The transformation step before the last one uses the equality (1). It is what has to be proved. </pre> <H3>Problem 9</H3>One solution of z^3+ A*z^2 - z - 14 = 0 is z = -2-root(3)*i, where A is a real number. Find A and other two solutions. <B>Solution</B> <pre> Since the given equation is with real coefficients (given), its complex roots are conjugated complex numbers. Since one such root is {{{z[1]}}} = {{{-2-i*sqrt(3)}}} (given), the other root is {{{z[2]}}} = {{{-2+i*sqrt(3)}}}. Vieta's theorem says that the product of three roots of the given equation equals to the constant term with the opposite sign. So {{{z[1]*z[2]*z[3]}}} = 14. The product of the two complex conjugated numbers {{{-2-i*sqrt(3)}}} and {{{-2+i*sqrt(3)}}} is {{{(-2)^2 + (sqrt(3))^2}}} = 4 + 3 = 7. Therefore, the third root of the given equation is {{{z[3]}}} = {{{(14)/7}}} = 2. Now the coefficient A equals to the sum of the three roots with the opposite sign, due to the same Vieta's theorem A = {{{-(z[1] + z[2] + z[3])}}} = {{{-( (-2-i*sqrt(3)) + (-2+i*sqrt(3)) + 2)}}} = -(-2-2+2) = 2. <U>ANSWER</U>. A = 2. The other two roots are {{{-2+i*sqrt(3)}}} and 2. </pre> My other lessons on quadratic equations in this website are - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/proof-of-quadratic-by-completing-the-square.lesson>PROOF of quadratic formula by completing the square</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Learning-by-examples-on-HOW-TO-complete-the-square.lesson>HOW TO complete the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-solve-quadratic-equation-by-completing-the-square-Learning-by-examples.lesson>HOW TO solve quadratic equation by completing the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A> - 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