SAT Algebra Perfect Score

To get a perfect score on SAT Math, you need to:

• Get every single problem correct.
• Have complete mastery of all of the SAT skills
• Remember the Tips and use them
• Figure out your common mistakes and avoid them

If $P^2=\sqrt{76+10\sqrt{3}}+5\sqrt{52-14\sqrt{3}},$ what is the negative number $P?$

(A) $\ \ -4$
(B) $\ \ -5$
(C) $\ \ -5.5$
(D) $\ \ -6$
(E) $\ \ -7$

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Correct Answer: D

Solution:

Using the technique of completing the square, we can rewrite the given equation as follows: $\begin{aligned} P^2 &=\sqrt{76+10\sqrt{3}}+5\sqrt{52-14\sqrt{3}}\\ &=\sqrt{(1+5\sqrt{3})^2}+5\sqrt{(7-\sqrt{3})^2}\\ &=(1+5\sqrt{3})+5(7-\sqrt{3})\\ &=36\\ \Rightarrow P&=\pm6. \end{aligned}$ Since $P$ is negative, $P=-6.$

Therefore, the correct answer is (D).

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Incorrect Choices:

(A), (B), (C) and (E)
The solution explains why these choices are wrong.

$a, b$ and $c$ are real numbers such that $\begin{array}{c}&a+b+c=2, &2^a+2^b+2^c=\frac{19}{2}, &2^{-a}+2^{-b}+2^{-c}=\frac{25}{8}.\end{array}$ What is the value of $4^a+4^b+4^c?$

(A) $\ \ \frac{257}{4}$
(B) $\ \ \frac{261}{4}$
(C) $\ \ \frac{265}{4}$
(D) $\ \ \frac{269}{4}$
(E) $\ \ \frac{273}{4}$

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Correct Answer: B

Solution:

We rewrite $4^a+4^b+4^c$ as follows to obtain: $\begin{aligned} 4^a+4^b+4^c &=\left(2^a\right)^2+\left(2^b\right)^2+\left(2^c\right)^2\\ &=\left(2^a+2^b+2^c\right)^2-2\left(2^a2^b+2^b2^c+2^c2^a\right)\\ &=\left(2^a+2^b+2^c\right)^2-\left(2^{a+b+1}+2^{b+c+1}+2^{c+a+1}\right)\\ &=\left(2^a+2^b+2^c\right)^2-\left(2^{3-c}+2^{3-a}+2^{3-b}\right)\\ &=\left(2^a+2^b+2^c\right)^2-2^3\cdot \left(2^{-c}+2^{-a}+2^{-b}\right)\\ &=\left(\frac{19}{2}\right)^2-8\cdot \frac{25}{8}=\frac{361-100}{4}=\frac{261}{4}. \end{aligned}$

Therefore, the correct answer is (B).

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Incorrect Choices:

(A), (C), (D) and (E)
The solution explains why these choices are wrong.

$x, y$ and $z$ are real numbers such that $\begin{array}{c}&2x+y+z=10, &2x^2-yz=-14.\end{array}$ If $m$ and $M$ are the minimum and maximum values of $xy+yz+zx,$ respectively, what is $M-m?$

(A) $\ \ 100$
(B) $\ \ 110$
(C) $\ \ 120$
(D) $\ \ 130$
(E) $\ \ 140$

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Correct Answer: C

Solution:

Observe that $y+z=-2x+10=-2(x-5)$ and $yz=2x^2+14.$ Then $y$ and $z$ are the two real roots of the following quadratic equation in $t:$ $t^2+2(x-5)+(2x^2+14).$ Since this equation has real roots, it must be true that its discriminant is non-negative: $\begin{aligned} \frac{D}{4}=(x-5)^2-(2x^2+14)=-x^2-10x+11 &\ge 0 \\ \Rightarrow x^2+10x-11&\le 0 \\ (x+11)(x-1)&\le 0\\ -11&\le x \le 1. \qquad (1) \end{aligned}$ Now, $\begin{aligned} xy+yz+zx &=yz+x(y+z)\\ &=(2x^2+14)+x\left(-2(x-5)\right)\\ &=10x+14. \qquad (2) \end{aligned}$ Since $-11\le x \le 1$ from $(1),$ the range of $(2)$ is $-96\le 10x+14 \le 24.$ Hence $m=-96$ and $M=24,$ which implies $M-m=24-(-96)=120.$

Therefore, the correct answer is (C).

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Incorrect Choices:

(A), (B), (D) and (E)
The solution explains why these choices are wrong.

Algebraic Manipulations

• Follow order of operations.

Polynomials

• $a^{2}-b^{2}=(a-b)(a+b)$
• $(a \pm b)^{2} = a^{2} \pm 2ab + b^{2}$

Exponents

• Know the Rules of Exponents.
• Recognize first few perfect squares (1, 4, 9, ... 400) and cubes (1, 8, 27,... 1000).
• The square of a number is always positive.
• $\sqrt{x^{2}} = \begin{cases} -x &\mbox{if } x < 0 \\ x & \mbox{if } x \geq 0. \\ \end{cases}$

Change the Subject

• Know the Rules of Exponents.

Inequalities

• $x^{2} \geq 0.$
• Know the Properties of Inequality.
• Multiplying (or dividing) both sides of an inequality by a negative number reverses its sign.
• Know the properties of numbers between $0$ and $1$.

Absolute Value

• $|x| = \begin{cases} -x &\mbox{if } x < 0 \\ x & \mbox{if } x \geq 0. \\ \end{cases}$
• $x^{2} \geq 0.$
• Know the Properties of Inequality.
• Multiplying (or dividing) both sides of an inequality by a negative number reverses its sign.

SAT General Tips