But I can't factor it like the previous one!

Algebra Level 3

$(x+3)(x+4)(x+6)(x+7) = 1120 + 1$

Find the sum of all real $x$ satisfying the equation above.

Inspiration

-10 10.5 -10.5 10

5 solutions

Chew-Seong Cheong
Mar 30, 2015

Let $y = x+5$ , then:

$\begin{aligned} (x+3)(x+4)(x+6)(x+7) & = 1121 \\ (y-1)(y-2) (y+1)(y+2) & = 1121 \\ (y^2-1)(y^2-4) & =1121 \\ y^4 - 5y^2 + 4 &= 1121 \\ y^4 - 5y^2 -1117 &= 0 \\ y^2 & = \dfrac{5 \pm \sqrt{4493}}{2} \\ \Rightarrow (x+5)^2 & =\dfrac{5 \pm \sqrt{4493}}{2} \\ \Rightarrow x^2+10x + 25 - \dfrac{5 + \sqrt{4493}}{2} & = 0 \end{aligned}$

Using Vieta's formula, the sum of $x$ satisfying the equation $= \boxed{-10}$

Actually, you haven't shown the main part! You should show that only one of the two equations produce real roots. This can be done by examination of approximations of $\Delta$ values of the two equations. We will get non-negative $\Delta$ value for only one equation out of the two and hence we conclude by Vieta's formula that the sum is $\boxed{-10}$ .

Note: For those who don't know, $\Delta$ here refers to the discriminant of a quadratic equation.

Prasun Biswas - 6 years, 2 months ago

You are right. I have edited the solution. When I solved the problem, I actually calculate out the actual values of the two $x$ . So I have eliminated the two other roots because of negative square roots. Then I noticed there is an easier way to explain this without realising the negative determinant part.

Chew-Seong Cheong - 6 years, 2 months ago

$(x+3)(x+4)(x+6)(x+7)=(x^2+7x+12)(x^2+13x+42)=(x^4+13x^3+42x^2+7x^3+91x^2+294x+12x^2+156x+504)=(x^4+13x^3+7x^3+42x^2+91x^2+12x^2+294x+156x+504)=x^4+20x^3+145x^2+450x+504$ .

Then, maybe hard to solve this biquadratic equation.

. . - 3 months ago

Nguyen Thanh Long
Mar 30, 2015

$(x^2+10x+21)*(x^2+10x+24)=1121$ $t=x^2+10x+21$ $t^2+3t-1121=0 (*)$ The equation (*) has one root that is: $t=\frac{\sqrt{4493}-3}{2}$ Solving the eq: $x^2+10x+21-\frac{\sqrt{4493}-3}{2}=0$ Get two roots for the equation that have sum is: $Sum=\boxed{10}$

sum = $-10$ , not $10$

Parth Lohomi - 6 years, 2 months ago

You are true. @Nguyen Thanh Long .

. . - 3 months ago

Daniel Rabelo
Apr 16, 2015

Let's do $x=v-5; v²=u+2.5$ .. Then we have: $(x+3)(x+4)(x+6)(x+7)=1121$

$(v-2)(v-1)(v+1)(v+2)=1121$

$(v^{2}-1)(v^{2}-4)=1121$

$(u+1.5)(u-1.5)=1121$

$u^{2}-2.25=1121$

$u=+-\sqrt{2.25+1121}$

As $v^2=u+2.5>0$ , $u=+\sqrt{2.25+1121}$ .

$v=+-\sqrt{2.5+\sqrt{2.25+1121}}$ and

$x=-5+-\sqrt{2.5+\sqrt{2.25+1121}}$

The sum: $x_1+x_2=-5+\sqrt{2.5+\sqrt{2.25+1121}}-5-\sqrt{2.5+\sqrt{2.25+1121}}$ $=-5-5=-10$

ps: $x=+-a$ is my notation for x=a or x=-a.

Use \pm or \mp command in LaTeX.

. . - 3 months ago

Then it appears like $\pm$ , and $\mp$ .

. . - 3 months ago

Jae Won Shin
Apr 23, 2015

Let $x+5=y$ then:

$(y-2)(y-1)(y+1)(y+2) =1121$

$(y^2-1)(y^2-4)=1121$

$y^{4}-5y^{2}+4=1121$

$y^{4}-5y^{2}-1117=0$

In here, according to Vieta's formula, no matter what the c value is in $ax^{2}+bx+c=0$ , the sum of two roots for x is always $\frac {-b}{a}$ . Hence in this case, the sum of roots for $y^{2}=5$ .

Substituting again for x,

$(x+5)^{2}=5$

$x^{2}+10x+25=5$

$x^{2}+10x+20=0$

*Note that we can still substitute $y^{2}$ to $(x+5)^{2}$ despite that it is an equal sign for their sum of roots, because it is also the two roots of x that's been put in the quadratic equation to add to that number (5).

Hence, using the Vieta's formula again, the summation of the two values of x is $\frac{-10}{1}= \boxed{-10}$

Jihoon Kang
Mar 30, 2015

There is only 1 x>0 which satisfies the equation. Then -x-10 is the only other solution where x<0 , then we get -(x+3) ×-(x+4)×-(x+6)×-(x+7)=(x+3)(x+4)(x+6)(x+7)=1120+1. So x+(-x-10)=-10

But I can't factor it like the previous one!

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