Problems using Vieta's formulas: Problems with Solutions

Solution:
The discriminant is [tex]D=(-7)^2-4\times 1\times(-44)=49+176=225>0[/tex], so the equation has two distinct real roots.
By Vieta's formulas:
[tex]x_1+x_2=-\frac{b}{a}=7[/tex]
[tex]x_1x_2=\frac{c}{a}=-44[/tex]
Check: the roots are [tex]11[/tex] and [tex]-4[/tex]; indeed [tex]11+(-4)=7[/tex] and [tex]11\times(-4)=-44[/tex].

Solution:
By the converse of Vieta's theorem, the numbers [tex]x_1[/tex] and [tex]x_2[/tex] are the roots of [tex]x^2-(x_1+x_2)x+x_1x_2=0[/tex].
[tex]x_1+x_2=-8+5=-3[/tex]
[tex]x_1x_2=-8\times 5=-40[/tex]
So the equation is [tex]x^2+3x-40=0[/tex].

Solution:
From Vieta's formulas we have [tex]-(x_1+x_2)=-54[/tex], so [tex]x_1+x_2=54[/tex]. We substitute [tex]x_1[/tex] for its value and get [tex]2+x_2=54[/tex], so [tex]x_2=52[/tex].

Solution:
If two numbers have sum [tex]24[/tex] and product [tex]128[/tex], then by the converse of Vieta's theorem they are the roots of
[tex]t^2-24t+128=0[/tex].
[tex]D=24^2-4\times 128=576-512=64[/tex], [tex]\sqrt{D}=8[/tex]
[tex]t_{1,2}=\frac{24\pm 8}{2}[/tex], so [tex]t_1=16[/tex], [tex]t_2=8[/tex].
The numbers are [tex]8[/tex] and [tex]16[/tex].

Solution:
By Vieta's formulas, the roots must satisfy
[tex]x_1+x_2=2m-1[/tex] and [tex]x_1x_2=m^2-m=m(m-1)[/tex].
The numbers [tex]m[/tex] and [tex]m-1[/tex] fit both conditions: [tex]m+(m-1)=2m-1[/tex] and [tex]m(m-1)=m^2-m[/tex].
By the converse of Vieta's theorem, the roots are [tex]x_1=m[/tex] and [tex]x_2=m-1[/tex].

Solution:
[tex]D=6^2-4\times 3\times(-16)=36+192=228>0[/tex], so there are two distinct real roots.
[tex]x_1x_2=\frac{c}{a}=-\frac{16}{3}<0[/tex], so the roots have opposite signs.
[tex]x_1+x_2=-\frac{b}{a}=-2<0[/tex], so the negative root is larger in absolute value.

Solution:
Expand and group:
[tex]x_1(x_2-2)+x_2(x_1-2)=2x_1x_2-2(x_1+x_2)[/tex].
[tex]D=9+20=29>0[/tex], and by Vieta's formulas [tex]x_1+x_2=3[/tex], [tex]x_1x_2=-5[/tex].
So the value is [tex]2\times(-5)-2\times 3=-10-6=-16[/tex].

Solution:
Let [tex]x_2=5x_1[/tex]. By Vieta's formulas:
[tex]x_1+x_2=-\frac{36}{3}=-12[/tex], so [tex]6x_1=-12[/tex], [tex]x_1=-2[/tex], [tex]x_2=-10[/tex].
[tex]x_1x_2=-\frac{m}{3}[/tex], so [tex](-2)\times(-10)=20=-\frac{m}{3}[/tex].
Hence [tex]m=-60[/tex].

Solution:
By Vieta's formulas, [tex]x_1+x_2=9[/tex], so [tex]x_2=9-x_1[/tex].
Substitute: [tex]4x_1+3(9-x_1)=34[/tex], so [tex]x_1+27=34[/tex], [tex]x_1=7[/tex] and [tex]x_2=2[/tex].
Then [tex]q=x_1x_2=7\times 2=14[/tex].

Solution:
By Vieta's formulas, [tex]x_1+x_2=8[/tex], so the other root is [tex]x_2=8-(4-\sqrt{3})=4+\sqrt{3}[/tex].
Then [tex]x_1x_2=(4-\sqrt{3})(4+\sqrt{3})=16-3=13[/tex].
Since [tex]x_1x_2=13c[/tex], we get [tex]13c=13[/tex], so [tex]c=1[/tex].

Solution:
The roots have opposite signs exactly when their product is negative: [tex]x_1x_2=2k+3<0[/tex], i.e. [tex]k<-\frac{3}{2}[/tex].
No separate check of the discriminant is needed: if [tex]\frac{c}{a}<0[/tex], then [tex]-4ac>0[/tex] and [tex]D=b^2-4ac>0[/tex] automatically.

Solution:
Let [tex]x=-3+\sqrt{2}[/tex]. Then [tex]x+3=\sqrt{2}[/tex]; squaring gives [tex]x^2+6x+9=2[/tex], i.e. [tex]x^2+6x+7=0[/tex].
Check with Vieta's formulas: the other root is [tex]-3-\sqrt{2}[/tex];
[tex](-3+\sqrt{2})+(-3-\sqrt{2})=-6=-p[/tex]
[tex](-3+\sqrt{2})(-3-\sqrt{2})=9-2=7=q[/tex]
Answer: [tex]x^2+6x+7=0[/tex].

Solution:
[tex]D=25+96=121>0[/tex]. By Vieta's formulas, [tex]x_1+x_2=\frac{5}{4}[/tex] and [tex]x_1x_2=\frac{6}{-4}=-\frac{3}{2}[/tex].
The new roots are [tex]y_1=2x_1[/tex], [tex]y_2=2x_2[/tex]:
[tex]y_1+y_2=2(x_1+x_2)=\frac{5}{2}[/tex]
[tex]y_1y_2=4x_1x_2=-6[/tex]
So [tex]y^2-\frac{5}{2}y-6=0[/tex], or after multiplying by [tex]2[/tex]: [tex]2y^2-5y-12=0[/tex].

Solution:
Identify the coefficients from the given equation. The quadratic equation is $2x^{2}+5x+3=0$.
By comparing it to the standard form $ax^{2}+bx+c=0$, we get: $a=2$ $b=5$ $c=3$
Use Vieta's formulas to find the sum and product of the roots.

Sum of roots: $\alpha +\beta =-\frac{b}{a}=-\frac{5}{2}$

Product of roots: $\alpha \beta =\frac{c}{a}=\frac{3}{2}$



Express the desired value in terms of the sum and product of the roots. We want to find the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$. Combine the fractions by finding a common denominator, which is $\alpha \beta: \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }$

Use the algebraic identity $(\alpha +\beta )^{2}=\alpha ^{2}+2\alpha \beta +\beta ^{2}$ to rewrite the numerator:$\alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}-2\alpha \beta $

Substitute this back into the expression:$\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }=\frac{(\alpha +\beta )^{2}-2\alpha \beta }{\alpha \beta }$

Substitute the values and calculate the final result. Substitute the sum $\alpha +\beta $ and product $\alpha \beta $ into the rewritten expression:$\frac{(-\frac{5}{2})^{2}-2(\frac{3}{2})}{\frac{3}{2}}$

Simplify the numerator: $\frac{\frac{25}{4}-\frac{6}{2}}{\frac{3}{2}}=\frac{\frac{25}{4}-\frac{12}{4}}{\frac{3}{2}}=\frac{\frac{13}{4}}{\frac{3}{2}}$

To divide by a fraction, multiply by its reciprocal: $\frac{13}{4}\times \frac{2}{3}=\frac{26}{12}$

Reduce the fraction: $\frac{26}{12}=\frac{13}{6}$

Solution:
Given Equation: $x^2 – 10x + 5 = 0$

By Vieta’s Formula
$\alpha + \beta = -\frac{b}{a} = -\frac{-10}{1} = 10$

$\alpha\beta = \frac{c}{a} = \frac{5}{1} = 5$

As $\alpha^2+\beta^2 = (\alpha + \beta )^2 - 2\alpha\beta= 10^2- 2\cdot5= 100 - 10$

$\alpha^2+\beta^2 = 90$

Now the value of $\frac{\alpha^2 + \beta^2}{\alpha^2\beta + \alpha\beta^2}= \frac{\alpha^2 + \beta^2}{\alpha\beta(\alpha + \beta)}= \frac{90}{5\times 10}= \frac{90}{50}= 1.8$