Pure Mathematics 1

Mathematics9709

Pure Mathematics 1

Mathematics9709

Quadratics in Another Form

A hidden quadratic contains one object and that same object squared. Replace the object, solve, then substitute back.

Spotting the pattern

Find the repeated object: in $4x^4-37x^2+9=0$ , it is $x^2$ because $x^4=(x^2)^2$ .
Let that object be $u$ , solve the quadratic in $u$ , then return to $x$ .
Apply the original restrictions after substituting back.

x^4,x^2

→

u=x^2

x^6,x^3

→

u=x^3

x,\sqrt{x}

→

u=\sqrt{x}

9^x,3^x

→

u=3^x

Higher powers

Worked example

Solve $4x^4-37x^2+9=0$

Let $u=x^2$ . Then $x^4=u^2$ , and the equation becomes a quadratic in $u$ .
$4u^2-37u+9=0$
Factorise and solve for $u$ .
$(4u-1)(u-9)=0$
$u=\tfrac{1}{4} \quad\text{or}\quad u=9$
Go back to $x$ . Each value of $u$ gives an equation $x^2=u$ . For positive $u$ , this equation has the two branches $x=\pm\sqrt u$ .
$x^2=\tfrac{1}{4} \;\Rightarrow\; x=\pm\tfrac{1}{2}$
$x^2=9 \;\Rightarrow\; x=\pm3$

Answer

x=\pm\tfrac{1}{2} \quad\text{or}\quad x=\pm3

For $u=x^2$ , keep both square-root signs. For $u=x^3$ , each real $u$ gives one cube root. For example, let $u=x^3$ in $8x^6+215x^3-27=0$ (9709/12/M/J/23 Q4):

8u^2+215u-27=0

(8u-1)(u+27)=0

u=\tfrac18\quad\text{or}\quad u=-27

x=\tfrac12\quad\text{or}\quad x=-3

Square roots

Worked example

Solve $x-4\sqrt{x}-12=0$

Let $u=\sqrt{x}$ . Then $x=u^2$ .
$u^2-4u-12=0$
Factorise and solve.
$(u-6)(u+2)=0$
$u=6 \quad\text{or}\quad u=-2$
Reject $u=-2$ , because $\sqrt{x}$ is never negative. Then square the root you keep.
$\sqrt{x}=6 \;\Rightarrow\; x=36$

Answer

x=36

Common mistake

u=6

means

\sqrt{x}=6

, not

x=6

. Substitute back before answering.

Powers of a number

Worked example

Solve $3(9^x)-28(3^x)+9=0$

Since $9=3^2$ , $9^x=3^{2x}=(3^x)^2$ . Let $u=3^x$ .
$3u^2-28u+9=0$
$(3u-1)(u-9)=0$
$u=\tfrac{1}{3} \quad\text{or}\quad u=9$
$3^x=3^{-1} \;\Rightarrow\; x=-1$
$3^x=3^2 \;\Rightarrow\; x=2$

Answer

x=-1 \quad\text{or}\quad x=2

Fractions that hide a quadratic

Worked example

Solve $6\sin\theta=1+\dfrac{2}{\sin\theta}$ for $-180^\circ<\theta<180^\circ$ (9709/11/M/J/25 Q1)

Require $\sin\theta\ne0$ , clear the fraction and rearrange.
$6\sin^2\theta-\sin\theta-2=0$
Let $s=\sin\theta$ and factorise.
$(2s+1)(3s-2)=0$
$\sin\theta=-\tfrac{1}{2} \quad\text{or}\quad \sin\theta=\tfrac{2}{3}$
Use the calculator's inverse-sine key to find one angle $\alpha$ . The other sine angle is $180^\circ-\alpha$ ; add or subtract $360^\circ$ until it lies in the stated interval.
$\sin\theta=-\tfrac12$
$\theta=-30^\circ,\ -150^\circ$
$\sin\theta=\tfrac23$
$\theta=41.8^\circ,\ 138.2^\circ$

Answer

\theta=-150^\circ,\ -30^\circ

\theta=41.8^\circ,\ 138.2^\circ

Your turnIndependent practice

Solve $x^4-5x^2+4=0$ for real $x$ .

Show worked answer

Let $u=x^2$ . Then solve $u^2-5u+4=0$ .

(u-1)(u-4)=0

u=1 \quad\text{or}\quad u=4

Return to $x$ and keep both square-root signs.

Answer

x=-2,\,-1,\,1,\,2

PreviousThe Quadratic Formula NextThe Graph and the Vertex

Quadratics in Another Form

A hidden quadratic contains one object and that same object squared. Replace the object, solve, then substitute back.

Spotting the pattern

Find the repeated object: in $4x^4-37x^2+9=0$ , it is $x^2$ because $x^4=(x^2)^2$ .
Let that object be $u$ , solve the quadratic in $u$ , then return to $x$ .
Apply the original restrictions after substituting back.

x^4,x^2

→

u=x^2

x^6,x^3

→

u=x^3

x,\sqrt{x}

→

u=\sqrt{x}

9^x,3^x

→

u=3^x

Higher powers

Worked example

Solve $4x^4-37x^2+9=0$

Let $u=x^2$ . Then $x^4=u^2$ , and the equation becomes a quadratic in $u$ .
$4u^2-37u+9=0$
Factorise and solve for $u$ .
$(4u-1)(u-9)=0$
$u=\tfrac{1}{4} \quad\text{or}\quad u=9$
Go back to $x$ . Each value of $u$ gives an equation $x^2=u$ . For positive $u$ , this equation has the two branches $x=\pm\sqrt u$ .
$x^2=\tfrac{1}{4} \;\Rightarrow\; x=\pm\tfrac{1}{2}$
$x^2=9 \;\Rightarrow\; x=\pm3$

Answer

x=\pm\tfrac{1}{2} \quad\text{or}\quad x=\pm3

For $u=x^2$ , keep both square-root signs. For $u=x^3$ , each real $u$ gives one cube root. For example, let $u=x^3$ in $8x^6+215x^3-27=0$ (9709/12/M/J/23 Q4):

8u^2+215u-27=0

(8u-1)(u+27)=0

u=\tfrac18\quad\text{or}\quad u=-27

x=\tfrac12\quad\text{or}\quad x=-3

Square roots

Worked example

Solve $x-4\sqrt{x}-12=0$

Let $u=\sqrt{x}$ . Then $x=u^2$ .
$u^2-4u-12=0$
Factorise and solve.
$(u-6)(u+2)=0$
$u=6 \quad\text{or}\quad u=-2$
Reject $u=-2$ , because $\sqrt{x}$ is never negative. Then square the root you keep.
$\sqrt{x}=6 \;\Rightarrow\; x=36$

Answer

x=36

Common mistake

u=6

means

\sqrt{x}=6

, not

x=6

. Substitute back before answering.

Powers of a number

Worked example

Solve $3(9^x)-28(3^x)+9=0$

Since $9=3^2$ , $9^x=3^{2x}=(3^x)^2$ . Let $u=3^x$ .
$3u^2-28u+9=0$
$(3u-1)(u-9)=0$
$u=\tfrac{1}{3} \quad\text{or}\quad u=9$
$3^x=3^{-1} \;\Rightarrow\; x=-1$
$3^x=3^2 \;\Rightarrow\; x=2$

Answer

x=-1 \quad\text{or}\quad x=2

Fractions that hide a quadratic

Worked example

Solve $6\sin\theta=1+\dfrac{2}{\sin\theta}$ for $-180^\circ<\theta<180^\circ$ (9709/11/M/J/25 Q1)

Require $\sin\theta\ne0$ , clear the fraction and rearrange.
$6\sin^2\theta-\sin\theta-2=0$
Let $s=\sin\theta$ and factorise.
$(2s+1)(3s-2)=0$
$\sin\theta=-\tfrac{1}{2} \quad\text{or}\quad \sin\theta=\tfrac{2}{3}$
Use the calculator's inverse-sine key to find one angle $\alpha$ . The other sine angle is $180^\circ-\alpha$ ; add or subtract $360^\circ$ until it lies in the stated interval.
$\sin\theta=-\tfrac12$
$\theta=-30^\circ,\ -150^\circ$
$\sin\theta=\tfrac23$
$\theta=41.8^\circ,\ 138.2^\circ$

Answer

\theta=-150^\circ,\ -30^\circ

\theta=41.8^\circ,\ 138.2^\circ

Your turnIndependent practice

Solve $x^4-5x^2+4=0$ for real $x$ .

Show worked answer

Let $u=x^2$ . Then solve $u^2-5u+4=0$ .

(u-1)(u-4)=0

u=1 \quad\text{or}\quad u=4

Return to $x$ and keep both square-root signs.

Answer

x=-2,\,-1,\,1,\,2

PreviousThe Quadratic Formula NextThe Graph and the Vertex